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NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combinations

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NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combinations

Edited By Ramraj Saini | Updated on Sep 23, 2023 06:09 PM IST

Permutation and Combinations Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combinations are provided here. Permutation means an arrangement where order matters and a combination means choosing a group of items where order doesn't matter. In this NCERT chapter, you will get NCERT solutions for class 11 maths chapter 7 permutation and combinations. These NCERT Solutions are prepared keeping in mind latest CBSE syllabus 2023 by the Careers360 subjects experts. Students can practice these solutions to get a good hold in the concepts. Here you will get NCERT solutions for class 11 also.

Let's consider 6 pants and 3 shirts, how many pairs you can wear? Suppose the pants are represented by p1,p2,p3,p4,p5,p6 and shirt by s1, s2, s3 then the possible pairs are (p1,s1), (p1,s2), (p1,s3), (p2,s1), (p2,s2), (p2,s3), (p3,s1), (p3,s2), (p3,s3), (p4,s1), (p4,s2), (p4,s3), (p5,s1), (p5,s2), (p5,s3), (p6,s1), (p6,s2), (p6,s3). So 18 pairs can be made. If this number is much higher, then how will you proceed to solve this type of problem? In solutions of NCERT for class 11 maths chapter 7 permutation and combinations, you will learn to solve this type of problem in a systematic manner. Without listing the pairs we can simply say that 6\times 3 pairs are possible. This chapter 7 class 11 maths starts with the fundamental principle of counting and then moves to the concepts of permutation and combinations. These concepts ch 7 maths class 11 are useful in solving many real-life problems related to selection and arrangement.

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Permutation and Combinations Class 11 Questions And Answers PDF Free Download

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Permutation and Combinations Class 11 Solutions - Important Formulae

Factorial:

Factorial is the continued product of the first n natural numbers and is denoted by n!.

The formula for factorial is n! = n(n – 1)(n – 2)… 3 × 2 × 1.

Special cases are 0! = 1 and 1! = 1.

Permutations:

Permutations refer to the various arrangements that can be constructed from a set of things.

The number of arrangements of n objects taken r at a time (where 0 < r ≤ n) is denoted by nPr.

The formula for permutations is nPr = n! / (n−r)!.

Permutations with Repeated Elements:

When dealing with permutations of n objects, where p1 are of one kind, p2 are of the second kind, and so on (such that p1 + p2 + p3 + … + pk = n), the formula for permutations is n! / (p1! p2! p3! ….. pk!).

Combinations:

Combinations are selections formed by taking some or all of a number of objects, regardless of their arrangement.

The number of r objects chosen from a set of n objects is indicated by nCr.

The formula for combinations is nCr = n! / (r!(n−r)!).

Relation Between Permutations and Combinations:

The relationship between permutations and combinations is given by two theorems:

  • nPr = nCr * r! when 0 < r ≤ n.
  • nCr + nCr-1 = (n+1)Cr.
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Permutation and Combinations Class 11 NCERT Solutions (Intext Questions and Exercise)

Permutation and combinations class 11 questions and answers - Exercise: 7.1

Question:1(i) How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
repetition of the digits is allowed?

Answer:

The five digits are 1, 2, 3, 4 and 5

As we know that repetition of the digits is allowed,

so , unit place can be filled by any of five digits.

Similarly , tens and hundreds digits can also be filled by any of five digits.

\therefore Number of 3-digit numbers can be formed =5\times 5\times 5=125

Question:1(ii) How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
repetition of the digits is not allowed?

Answer:

The five digits are 1, 2, 3, 4 and 5

As we know that repetition of the digits is not allowed,

so, the unit place can be filled by any of five digits.

Tens place can be filled with any of the remaining four digits.

Hundreds place can be filled with any of the remaining three digits.

\therefore Number of 3-digit numbers can be formed when repetition is not allowed =5\times 4\times 3=60

Question:2 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Answer:

The six digits are 1, 2, 3, 4 ,5 and 6

As we know that repetition of the digits is allowed,

so, the unit place can be filled by any of even digits i.e.2,4 or 6

Similarly, tens and hundreds of digits can also be filled by any of six digits.

\therefore Number of 3-digit even numbers can be formed =3\times 6\times 6=108

Question:3 How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Answer:

There are 10 letters of English alphabets.

As we know that repetition of the letters is not allowed,

so, the first place can be filled by any of 10 letters.

Second place can be filled with any of the remaining 9 letters.

Third place can be filled with any of the remaining 8 letters.

The fourth place can be filled with any of the remaining 7 letters.

\therefore Number of 4-letter code can be formed when the repetition of letters is not allowed =10\times 9\times 8\times 7=5040

Hence, 5040 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated.

Question:4 How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Answer:

It is given that 5-digit telephone numbers always start with 67.

First, two digits among 5-digit telephone numbers are fixed and rest 3 digits are variable.

6,7,-,-,-

The 10 digits are from 0 to 9.

As we know that repetition of the digits is not allowed,

so, the first and second place is filled by two digits 67

Third place can be filled with any of the remaining 8 digits.

The fourth place can be filled with any of the remaining 7 digits.

The fifth place can be filled with any of the remaining 6 digits.

\therefore Number of 5-digit telephone numbers can be formed when repetition is not allowed =8\times 7\times 6=336

Question:5 A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Answer:

When a coin is tossed the number of outcomes is 2 i.e. head or tail.

When a coin is tossed 3 times then by multiplication principle,

the number of outcomes =2\times 2\times 2=8

Question:6 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Answer:

Each signal requires use of 2 flags.

There will be as many flags as there are ways of filling in 2 vacant places in succession by the given 5 flags of different colours.

The upper vacant place can be filled in 5 different ways with any of 5 flags and lower vacant place can be filled in 4 different ways by any of rest 4 flags.

Hence, by multiplication principle number of different signals that can be generated =5\times 4=20

Class 11 maths chapter 7 question answer - Exercise: 7.2

Question:1(i) Evaluate \; 8 !

Answer:

Factorial can be given as :

8!=8\times 7\times 6\times 5\times \times 4\times 3\times 2\times 1=40320

Question:1(ii) Evaluate \; 4!-3!

Answer:

Factorial can be given as :

(ii)\; 4!-3!

=( 4\times 3\times 2\times 1)-(3\times 2\times 1)

=24-6

=18

Question:2 Is 3!+4!=7!\; ?

Answer:

Factorial can be given as :

To prove : 3!+4!=7!

R.H.S : 3!+4!

=(3\times 2\times 1)+( 4\times 3\times 2\times 1)

=6+24

=30

L.H.S : 7!

=7\times 6\times 5\times 4\times 3\times 2\times 1=5040

L.H.S\neq R.H.S

Question:3 Compute \frac{8!}{6!\times 2!}

Answer:

To compute the factorial :

\frac{8!}{6!\times 2!}=\frac{8\times 7\times 6!}{6!\times 2\times 1}

=\frac{8\times 7}{ 2}

=4\times 7=28

So the answer is 28

Question:4 If \frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}, find x

Answer:

Factorial can be given as :

To find x : \frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!},

R.H.S :\, \, \frac{1}{6!}+\frac{1}{7!}

= \frac{1}{6!}+\frac{1}{7\times 6!}

= \frac{1}{6!}(1+\frac{1}{7})

= \frac{1}{6!}(\frac{8}{7})

L.H.S:\, \, \frac{x}{8!},

=\frac{x}{8\times 7\times 6!}

Given : L.H.S= R.H.S

\frac{1}{6!}\left ( \frac{8}{7} \right )=\frac{x}{8\times 7\times 6!}

\Rightarrow \, \, 8=\frac{x}{8}

\Rightarrow \, \, x=8\times 8=64

Question:5(i) Evaluate \frac{n!}{(n-r)!}, when

\; n=6,r=2

Answer:

To evaluate \frac{n!}{(n-r)!},

Put n=6,r=2

\frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}

=\frac{6!}{4!}

=\frac{6\times 5\times 4!}{4!}

=6\times 5=30

Question:5(ii) Evaluate \frac{n!}{(n-r)!}, when

\; n=9,r=5

Answer:

To evaluate \frac{n!}{(n-r)!},

Put n=9,r=5

\frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}

=\frac{9!}{4!}

=\frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}

=9\times 8\times 7\times 6\times 5=15120

Class 11 maths chapter 7 question answer - Exercise: 7.3

Question:1 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Answer:

3-digit numbers have to be formed by using the digits 1 to 9.

Here, the order of digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers =^{9}P_3

=\frac{9!}{(9-3)!}

=\frac{9!}{6!}

=\frac{9\times 8\times 7\times 6!}{6!}

=9\times 8\times 7=504

Question:2 How many 4-digit numbers are there with no digit repeated?

Answer:

The thousands place of 4-digit numbers has to be formed by using the digits 1 to 9(0 cannot be included).

Therefore, the number of ways in which thousands place can be filled is 9.

Hundreds,tens, unit place can be filled by any digits from 0 to 9.

The digit cannot be repeated in 4-digit numbers and thousand places is occupied with a digit.

Hundreds, tens, unit place can be filled by remaining any 9 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers =^{9}P_3

=\frac{9!}{(9-3)!}

=\frac{9!}{6!}

=\frac{9\times 8\times 7\times 6!}{6!}

=9\times 8\times 7=504

Thus, by multiplication principle, required 4 -digit numbers is 9\times 504=4536

Question:3 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Answer:

3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated.

The unit place can be filled in 3 ways by any digits from 2,4 or 6.

The digit cannot be repeated in 3-digit numbers and the unit place is occupied with a digit(2,4 or 6).

Hundreds, tens place can be filled by remaining any 5 digits.

Therefore, there will be as many 2-digit numbers as there are permutations of 5 different digits taken 2 at a time.

Therefore, the required number of 2-digit numbers =^{5}P_2

=\frac{5!}{(5-2)!}

=\frac{5!}{3!}

=\frac{ 5\times 4\times 3!}{3!}

=5\times 4=20

Thus, by multiplication principle, required 3 -digit numbers is 3\times 20=60

Question:4 Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,5 if no digit is repeated. How many of these will be even?

Answer:

4-digit numbers that can be formed using the digits 1, 2, 3, 4,5.

Therefore, there will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

Therefore, the required number of 4-digit numbers =^{5}P_4

=\frac{5!}{(5-4)!}

=\frac{5!}{1!}

= 5\times 4\times 3\times 2\times 1=120


4-digit even numbers can be made using the digits 1, 2, 3, 4, 5 if no digit is repeated.

The unit place can be filled in 2 ways by any digits from 2 or 4.

The digit cannot be repeated in 4-digit numbers and the unit place is occupied with a digit(2 or 4).

Thousands, hundreds, tens place can be filled by remaining any 4 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 4 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers =^{4}P_3

=\frac{4!}{(4-3)!}

=\frac{4!}{1!}

=4\times 3\times 2\times 1=24

Thus, by multiplication principle, required 4 -digit numbers is 2\times 24=48

Question:5 From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Answer:

From a committee of 8 persons, chairman and a vice chairman are to be chosen assuming one person can not hold more than one position.

Therefore,number of ways of choosing a chairman and a vice chairman is permutations of 8 different objects taken 2 at a time.

Therefore, required number of ways =^{8}P_2

=\frac{8!}{(8-2)!}

=\frac{8!}{6!}

=\frac{ 8\times 7\times 6!}{6!}

=8\times 7=56

Question:6 Find n if ^{n-1}P_{3}:^{n}P_{4}=1:9.

Answer:

Given : ^{n-1}P_{3}:^{n}P_{4}=1:9.

To find the value of n

^{n-1}P_{3}:^{n}P_{4}=1:9.

\Rightarrow \frac{^{n-1}P_3}{^nP_4}=\frac{1}{9}

\Rightarrow \, \, \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{n-4!}}=\frac{1}{9}

\Rightarrow \, \, \frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}

\Rightarrow \, \, \frac{(n-1)!}{n!}=\frac{1}{9}

\Rightarrow \, \, \frac{(n-1)!}{(n-1)!\times n}=\frac{1}{9}

\Rightarrow \, \, \frac{1}{ n}=\frac{1}{9}

\Rightarrow \, \, n=9.

Question:7(i) Find r if

^{5}P_{r}=2\; ^{6}\! P_{r -1}

Answer:

Given : ^{5}P_{r}=2\; ^{6}\! P_{r -1}

To find the value of r.

^{5}P_{r}=2\; ^{6}\! P_{r -1}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-(r-1))!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6\times 5!}{(6-r+1)!}

\Rightarrow \, \, \frac{1}{(5-r)!}=2\times \frac{6}{(7-r)!}

\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{12}{(7-r)\times (6-r)\times (5-r)!}

\Rightarrow \, \, \frac{1}{1}= \frac{12}{(7-r)\times (6-r)}

\Rightarrow \, \, (7-r)\times (6-r)=12

\Rightarrow \, \, 42-6r-7r+r^2=12

\Rightarrow \, \, r^2-13r+30=0

\Rightarrow \, \, r^2-3r-10r+30=0

\Rightarrow \, \, r(r-3)-10(r-3)=0

\Rightarrow \, \, (r-3)(r-10)=0

\Rightarrow \, \, r=3,10

We know that

^nP_r=\frac{n!}{(n-r)!}

where 0\leq r\leq n

\therefore 0\leq r\leq 5

Thus the value of, r=3

Question:7(ii) Find r if

^{5}P_{r}=^{6\! \! }P_{r-1}

Answer:

Given : ^{5}P_{r}=^{6\! \! }P_{r-1}

To find the value of r.

^{5}P_{r}=^{6\! \! }P_{r-1}

\Rightarrow \, \, \frac{5!}{(5-r)!}= \frac{6!}{(6-(r-1))!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6\times 5!}{(6-r+1)!}

\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)!}

\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)\times (6-r)\times (5-r)!}

\Rightarrow \, \, \frac{1}{1}= \frac{6}{(7-r)\times (6-r)}

\Rightarrow \, \, (7-r)\times (6-r)=6

\Rightarrow \, \, 42-6r-7r+r^2=6

\Rightarrow \, \, r^2-13r+36=0

\Rightarrow \, \, r^2-4r-9r+36=0

\Rightarrow \, \, r(r-4)-9(r-4)=0

\Rightarrow \, \, (r-4)(r-9)=0

\Rightarrow \, \, r=4,9

We know that

^nP_r=\frac{n!}{(n-r)!}

where 0\leq r\leq n

\therefore 0\leq r\leq 5

Thus, r=4

Question:8 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Answer:

There are 8 different letters in word EQUATION.

Therefore, words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once

is permutations of 8 different letters taken 8 at a time, which is ^8P_8=8!

Hence, the required number of words formed =8!=40320

Question:9(i) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

4 letters are used at a time

Answer:

There are 6 letters in word MONDAY.

Therefore, words that can be formed using 4 letters of the word MONDAY.

Hence, the required number of words formed using 4 letters =^6P_4

=\frac{6!}{(6-4)!}

=\frac{6!}{(2)!}

=\frac{6\times 5\times 4\times 3\times 2!}{(2)!}

=6\times 5\times 4\times 3

=360

Question:9(ii) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

all letters are used at a time

Answer:

There are 6 letters in word MONDAY.

Therefore, words that can be formed using all 6 letters of the word MONDAY.

Hence, the required number of words formed using 6 letters at a time =^6P_6

=\frac{6!}{(6-6)!}

=\frac{6!}{(0)!}

=6\times 5\times 4\times 3\times 2\times 1

=720

Question:9(iii) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

all letters are used but first letter is a vowel?

Answer:

There are 6 letters and 2 vowels in word MONDAY.

Therefore, the right most position can be filled by any of these 2 vowels in 2 ways.

Remaining 5 places of the word can be filled using any of rest 5 letters of the word MONDAY.

Hence, the required number of words formed using 5 letters at a time =^5P_5

=5!

=5\times 4\times 3\times 2\times 1

=120

Words formed starting from vowel using 6 letters =2\times 120=240

Question:10 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Answer:

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, M appears 1 time and P appear 2 times.

Therefore, the number of distinct permutations of letters of the given word is

=\frac{11!}{4!4!2!}

=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!4!2!}

=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 2}

=34650


There are 4 I's in the given word. When they occur together they are treated as a single object for the time being. This single object with the remaining 7 objects will be 8 objects.

These 8 objects in which there are 4Ss and 2Ps can be arranged in \frac{8!}{4!2!}=840 ways.

The number of arrangement where all I's occur together = 840.


Hence, the distinct permutations of the letters in MISSISSIPPI in which the four I's do not come together =34650-840=33810

Question:11(i) In how many ways can the letters of the word PERMUTATIONS be arranged if the

words start with P and end with S?

Answer:

There are 2 T's in word PERMUTATIONS, all other letters appear at once only.

If words start with P and ending with S i.e. P and S are fixed, then 10 letters are left.

The required number of arrangements are

=\frac{10!}{2!}=1814400

Question:11(ii) In how many ways can the letters of the word PERMUTATIONS be arranged if the

vowels are all together?

Answer:

There are 5 vowels in word PERMUTATIONS and each appears once.

Since all 5 vowels are to occur together so can be treated as 1 object.

The single object with the remaining 7 objects will be 8 objects.

The 8 objects in which 2 T's repeat can be arranged as

=\frac{8!}{2!} ways.

These 5 vowels can also be arranged in 5! ways.

Hence, using the multiplication principle, the required number of arrangements are

=\frac{8!}{2!}\times 5!=2419200 ways.

Question:11(iii) In how many ways can the letters of the word PERMUTATIONS be arranged if the

there are always 4 letters between P and S?

Answer:

The letters of the word PERMUTATIONS be arranged in such a way that there are always 4 letters between P and S.

Therefore, in a way P and S are fixed. The remaining 10 letters in which 2 T's are present can be arranged in

=\frac{10!}{2!} ways.

Also, P and S can be placed such that there are 4 letters between them in 2\times 7=14 ways.

Therefore, using the multiplication principle required arrangements

=\frac{10!}{2!}14=25401600

Permutation and combinations class 11 NCERT solutions - Exercise: 7.4

Question:1 If ^{n}C_{8}=^{n}\; \! \! \! \! C_{2}, find ^{n}C_{2}

Answer:

Given : ^{n}C_{8}=^{n}\; \! \! \! \! C_{2},

We know that ^nC_a=^nC_b \Rightarrow a=b\, \, \, or\, \, n=a+b

^{n}C_{8}=^{n}\; \! \! \! \! C_{2},

\Rightarrow n=8+2

\Rightarrow n=10


^{n}C_{2} =^{10}C_2

=\frac{10!}{(10-2)!2!}

=\frac{10!}{8!2!}

=\frac{10\times 9\times 8!}{8!2!}

=5\times 9=45

Thus the answer is 45

Question:2(i) Determine n if

\; ^{2n}C_{3}:^{n}C_{3}=12:1

Answer:

Given that : (i)\; ^{2n}C_{3}:^{n}C_{3}=12:1

\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{12}{1}

The ratio can be written as

\Rightarrow \, \, \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{12}{1}

\Rightarrow \, \, {\frac{2n!(n-3)!}{(2n-3)!n!}}=\frac{12}{1}

\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times(n-2)\times (n-3)! }=\frac{12}{1}

\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)}{ n\times (n-1)\times(n-2) }=\frac{12}{1}

\Rightarrow \, \, \frac{2\times (2n-1)\times 2}{ (n-2) }=\frac{12}{1}

\Rightarrow \, \, \frac{4\times (2n-1)}{ (n-2) }=\frac{12}{1}

\Rightarrow \, \, \frac{ (2n-1)}{ (n-2) }=\frac{3}{1}

\Rightarrow \, \, 2n-1=3n-6

\Rightarrow \, \, 6-1=3n-2n

\Rightarrow \, \, n=5

Question:2(ii) Determine n if

\; ^{2n}C_{3}:^{n}C_{3}=11:1

Answer:

Given that : (ii)\; ^{2n}C_{3}:^{n}C_{3}=11:1

\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{11}{1}

\Rightarrow \, \, \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{11}{1}

\Rightarrow \, \, {\frac{2n!(n-3)!}{(2n-3)!n!}}=\frac{11}{1}

\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times(n-2)\times (n-3)! }=\frac{11}{1}

\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)}{ n\times (n-1)\times(n-2) }=\frac{11}{1}

\Rightarrow \, \, \frac{2\times (2n-1)\times 2}{ (n-2) }=\frac{11}{1}

\Rightarrow \, \, \frac{4\times (2n-1)}{ (n-2) }=\frac{11}{1}

\Rightarrow \, \, 8n-4=11n-22

\Rightarrow \, \, 22-4=11n-8n

\Rightarrow \, \, 3n=18

\Rightarrow \, \, n=6

Thus the value of n=6

Question:3 How many chords can be drawn through 21 points on a circle?

Answer:

To draw chords 2 points are required on the circle.

To know the number of chords on the circle , when points on the circle are 21.

Combinations =Number of chords =^{21}C_2

=\frac{21!}{(21-2)!2!}

=\frac{21!}{19!2!}

=\frac{21\times 20}{2}

=210

Question:4 In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer:

A team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ^5C_3 ways.

3 girls can be selected from 4 boys in ^4C_3 ways.

Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected =^5C_3\times ^4C_3

=\frac{5!}{2!3!}\times \frac{4!}{1!3!}

=10\times 4=40

Question:5 Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer:

There are 6 red balls, 5 white balls and 5 blue balls.

9 balls have to be selected in such a way that consists of 3 balls of each colour.

3 balls are selected from 6 red balls in ^6C_3 .

3 balls are selected from 5 white balls in ^5C_3

3 balls are selected from 5 blue balls in ^5C_3 .

Hence, by the multiplication principle, the number of ways of selecting 9 balls =^6C_3\times ^5C_3\times ^5C_3

=\frac{6!}{3!3!}\times \frac{5!}{2!3!}\times \frac{5!}{2!3!}

=20\times 10\times 10=2000

Question:6 Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Answer:

In a deck, there is 4 ace out of 52 cards.

A combination of 5 cards is to be selected containing exactly one ace.

Then, one ace can be selected in ^4C_1 ways and other 4 cards can be selected in ^4^8C_4 ways.

Hence, using the multiplication principle, required the number of 5 card combination =^4C_1\times ^4^8C_4

=\frac{4!}{1!3!}\times \frac{48!}{44!4!}

=4\times \frac{48\times 47\times 46\times 45}{4\times 3\times 2}=778320

Question:7 In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer:

Out off, 17 players, 5 are bowlers.

A cricket team of 11 is to be selected such that there are exactly 4 bowlers.

4 bowlers can be selected in ^5C_4 ways and 7 players can be selected in ^1^2C_7 ways.

Thus, using multiplication priciple, number of ways of selecting the team =^5C_4 .^1^2C_7

=\frac{5!}{1!4!}\times \frac{12!}{5!7!}

=5 \times \frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}

=3960

Question:8 A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Answer:

A bag contains 5 black and 6 red balls.

2 black balls can be selected in ^5C_2 ways and 3 red balls can be selected in ^6C_3 ways.

Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls =^5C_2 .^6C_3

=\frac{5!}{2!3!}\times \frac{6!}{3!3!}

=10\times 20=200

Question:9 In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

9 courses are available and 2 specific courses are compulsory for every student.

Therefore, every student has to select 3 courses out of the remaining 7 courses.

This can be selected in ^7C_3 ways.

Thus, using multiplication priciple, number of ways of selecting courses =^7C_3

=\frac{7!}{3!4!}

= \frac{7\times 6\times 5}{ 3\times 2}

=35

Permutation and combinations class 11 solutions - Miscellaneous Exercise

Question:1 How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Answer:

In the word DAUGHTER, we have

vowels = 3(A,E,U)

consonants = 5(D,G,H,T,R)

Number of ways of selecting 2 vowels =^3C_2

Number of ways of selecting 3 consonants =^5C_3

Therefore, the number of ways of selecting 2 vowels and 3 consonants =^3C_2 .^5C_3

=3\times 10=30

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged in 5! ways.

Thus, the required number of different words = 5!\times 30=120\times 30=3600

Question:2 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Answer:

In the word EQUATION, we have

vowels = 5(A,E,I,O,U)

consonants = 3(Q,T,N)

Since all the vowels and consonants occur together so (AEIOU) and (QTN) can be assumed as single objects.

Then, permutations of these two objects taken at a time =^2P_2=2!=2

Corresponding to each of these permutations, there are 5! permutations for vowels and 3! permutations for consonants.

Thus, by multiplication principle, required the number of different words = 2\times 5!\times 3!=2\times 120\times 6=1440

Question:3(i) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

exactly 3 girls?

Answer:

There are 9 boys and 4 girls. A committee of 7 has to be formed.

Given : Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways =^4C_3.^9C_4

=\frac{4!}{3!1!}\times \frac{9!}{4!5!}

=4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}

= 9\times 8\times 7=504

Question:3(ii) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

at least 3 girls?

Answer:

There are 9 boys and 4 girls. A committee of 7 has to be formed.

(ii) at least 3 girls, there can be two cases :

(a) Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways =^4C_3.^9C_4

=\frac{4!}{3!1!}\times \frac{9!}{4!5!}

=4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}

= 9\times 8\times 7=504

(b) Girls =4, so boys in committee= 7-4=3

Thus, the required number of ways =^4C_4.^9C_3

=\frac{4!}{4!0!}\times \frac{9!}{3!6!}

= \frac{9\times 8\times 7\times 6!}{ 3\times 2\times 6!}

=84


Hence, in this case, the number of ways = 504+84=588

Question:3(iii) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

atmost 3 girls?

Answer:

There are 9 boys and 4 girls. A committee of 7 has to be formed.

(ii) atmost 3 girls, there can be 4 cases :

(a) Girls =0, so boys in committee= 7-0=7

Thus, the required number of ways =^4C_0.^9C_7

=\frac{4!}{4!0!}\times \frac{9!}{2!7!}

=9\times 4=36

(b) Girls =1, so boys in committee= 7-1=6

Thus, the required number of ways =^4C_1.^9C_6

=\frac{4!}{3!1!}\times \frac{9!}{3!6!}

=336

(c) Girls =2, so boys in committee= 7-2=5

Thus, the required number of ways =^4C_2.^9C_5

=\frac{4!}{2!2!}\times \frac{9!}{4!5!}

=756

(d) Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways =^4C_3.^9C_4

=\frac{4!}{3!1!}\times \frac{9!}{4!5!}

=504

Hence, in this case, the number of ways = 36+336+756+504=1632

Question:4 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Answer:

In the word EXAMINATION, we have 11 letters out of which A,I, N appear twice and all other letters appear once.

The word that will be listed before the first word starting with E will be words starting with A.

Therefore, to get the number of words starting with A, letter A is fixed at extreme left position, the remaining 10 letters can be arranged.

Since there are 2 I's and 2 N's in the remaining 10 letters.

Number of words starting with A =\frac{10!}{2!2!}=907200

Thus, the required number of different words = 907200

Question:5 How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Answer:

For the number to be divisible by 10, unit digit should be 0.

Thus, 0 is fixed at a unit place.

Therefore, the remaining 5 places should be filled with 1,3,5,7,9.

The remaining 5 vacant places can be filled in 5! ways.

Hence, the required number of 6 digit numbers which are divisible by 10 =5!=120

Question:6 The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Answer:

Two different vowels and 2 different consonants are to be selected from the English alphabets.

Since there are 5 different vowels so the number of ways of selecting two different vowels = ^5C_2

=\frac{5!}{2!3!}=10

Since there are 21 different consonants so the number of ways of selecting two different consonants = ^2^1C_2

=\frac{21!}{2!19!}=210

Therefore, the number of combinations of 2 vowels and 2 consonants =10\times 210=2100

Each of these 2100 combinations has 4 letters and these 4 letters arrange among themselves in 4! ways.

Hence, the required number of words =210\times 4!=50400

Question:7 In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Answer:

It is given that a question paper consists of 12 questions divided in two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively.

A student is required to attempt 8 questions in all, selecting at least 3 from each part.

This can be done as follows:

(i) 3 questions from part I and 5 questions from part II

(ii) 4 questions from part I and 4 questions from part II

(iii) 5 questions from part I and 3 questions from part II

3 questions from part I and 5 questions from part II can be selected in ^5C_3.^7C_5 ways.

4 questions from part I and 4 questions from part II can be selected in ^5C_4.^7C_4 ways.

5 questions from part I and 3 questions from part II can be selected in ^5C_5.^7C_3 ways.

Hence, required number of ways of selecting questions :

=^5C_3.^7C_5+^5C_4.^7C_4+^5C_5.^7C_3

=\frac{5!}{2!3!}\times \frac{7}{2!5!}+\frac{5!}{1!4!}\times \frac{7}{4!3!}+\frac{5!}{5!0!}\times \frac{7}{3!4!}

=210+175+35

=420

Question:8 Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Answer:

From a deck of 52 cards, 5 cards combinations have to be made in such a way that in each selection of 5 cards there is exactly 1 king.

Number of kings =4

Number of ways of selecting 1 king =^4C_1

4 cards from the remaining 48 cards are selected in ^4^8C_4 ways.

Thus, the required number of 5 card combinations =^4C_1 .^4^8C_4

Question:9 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Answer:

It is required to seat 5 men and 4 women in a row so that the women occupy the even places.
The 5 men can be seated in 5! ways.

4 women can be seated at cross marked places (so that women occupy even places)

Therefore, women can be seated in 4! ways.

Thus, the possible arrangements =5!\times 4!=120\times 24=2880

Question:10 From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?

Answer:

From a class of 25 students, 10 are to be chosen for an excursion party.

There are 3 students who decide that either all of them will join or none of them will join, there are two cases :

The case I: All 3 off them join.

Then, the remaining 7 students can be chosen from 22 students in ^2^2C_7 ways.

Case II : All 3 of them do not join.

Then,10 students can be chosen from 22 students in ^2^2C_1_0 ways.

Thus, the required number of ways for the excursion of party =^2^2C_7+^2^2C_1_0

Question:11 In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Answer:

In the word ASSASSINATION, we have

number of S =4

number of A =3

number of I= 2

number of N =2

Rest of letters appear at once.

Since all words have to be arranged in such a way that all the S are together so we can assume SSSS as an object.

The single object SSSS with other 9 objects is counted as 10.

These 10 objects can be arranged in (we have 3 A's,2 I's,2 N's)

=\frac{10!}{3!2!2!} ways.

Hence, requires a number of ways of arranging letters

=\frac{10!}{3!2!2!}=151200

Permutation And Combinations Class 11 Solutions - Topics

  • Introduction
  • Fundamental Principle of Counting
  • Permutations
    • Permutations when all objects are distinct
    • Factorial notation
    • Derivation of the formula for nPr
    • Permutations when all objects are not distinct
  • Combinations

Interested students can find class 11 maths ch 7 question answer of all exercises here.

Permutation and Combination Class 11 exercise 7.1 6 Questions

Permutation and Combination Class 11 exercise 7.2 5 Questions

Permutation and Combination Class 11 exercise 7.3 11 Questions

Permutation and Combination Class 11 exercise 7.4 9 Questions

Permutation and Combination Class 11 miscellaneous exercise 11 Questions

NCERT solutions for class 11- Chapter wise

Key feature of class 11 maths chapter 7 NCERT solutions

Prepared by Experts: The material for ch 7 maths class 11 solutions has been prepared by subject matter experts who have extensive experience in teaching Mathematics.

Prepared Keeping in Latest Syllabus of CBSE: The material for maths chapter 7 class 11 has been designed according to the latest syllabus prescribed by the Central Board of Secondary Education (CBSE) for Class 11 Maths.

Easy and Simple Language: The language used in class 11 chapter 7 material is easy to understand and simple, making it easy for students to grasp the concepts and apply them in problem-solving.

NCERT solutions for class 11- Subject wise

There are some important formulas in NCERT solutions for class 11 maths chapter 7 permutation and combinations which you should remember:

In simple words, a permutation is the number of possible arrangments within a fixed group where order matters and to calculate permutation you can use the formula mentioned below.

^{n} \mathrm{P}_{r}=\frac{n !}{(n-r) !}, 0 \leq r \leq n

^{n} \mathrm{P}_{r}=\frac{n(n-1)(n-2) \ldots(n-r+1)(n-r)(n-r-1) \ldots 3 \times 2 \times 1}{(n-r)(n-r-1) \ldots 3 \times 2 \times 1}

^{n} \mathrm{P}_{r}=^{n} \mathrm{C}_{r} \times r !, 0<r \leq n

In simple words, the combinations mean how many groups can be formed from a large number of people. Here you are not considering the order as you are just forming groups from a large group. To calculate the combinations you can use the formula mentioned below.

^{n} \mathrm{C}_{r}=\frac{n !}{r !(n-r) !}

^{n} C_{n-r}=\frac{n !}{(n-r) !(n-(n-r)) !}=\frac{n !}{(n-r) ! r !}=^{n} \mathbf{C}_{r}

In this chapter, there are 11 problems in miscellaneous exercise. To get command on this chapter you should solve miscellaneous exercise too. In permutation and combinations NCERT solutions, you will get solutions to miscellaneous exercise too. This chapter very important for the probability which you will be studying in class 11 and 12 both.

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Question (FAQs)

1. What are important topics of the chapter Permutation and Combinations ?

The fundamental principle of counting, permutations, and combinations are covered in this NCERT Book chapter. you can also go through NCERT Syllabus, where you can find important topics enumerated in this chapter permutation and combination class 11.

2. Discuss the fundamental concepts of Permutations and Combinations in the NCERT Solutions for Class 11 Maths Chapter 7.

the chapter permutation and combination solutions discuses fundamental concepts such as permutations refer to the various arrangements of a set of elements or numbers, whereas combinations represent a specific arrangement of distinct elements. The primary concepts that form the basis of these concepts are the fundamental theorem of counting, which states that if there are 'n' ways of one event occurring and 'm' ways of another event occurring, then the total number of occurrences can be found by multiplying n and m. The second concept pertains to the number of permutations of 'n' distinct elements taken 'r' at a time without repetition, which can be expressed as nPr. In cases where repetition is allowed, this is denoted as nr.

3. How many exercises are there in permutation and combination Class 11?

The permutation and combination class 11 ncert solutions contains four exercises and one  miscellaneous exercise which are listed below.

Permutation and Combination Class 11 exercise 7.1 6 Questions

Permutation and Combination Class 11 exercise 7.2 5 Questions

Permutation and Combination Class 11 exercise 7.3 11 Questions

Permutation and Combination Class 11 exercise 7.4 9 Questions

Permutation and Combination Class 11 miscellaneous exercise 11 Questions

4. Where can I find the complete solutions of NCERT for class 11 maths ?

Here students can get the detailed NCERT solutions for class 11 maths  by clicking on the link. After practicing these solutions students can get confidence on the concepts that lead to good score in the cbse exam. The chapter maths class 11 chapter 7 is also foundation for multiple chapters such as probability and statistics therefore students should practice. Students can study permutation and combination class 11 pdf both online and offline mode.

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Veterinary Doctor
5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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