NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.3 - Permutations and Combinations

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 the 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

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