NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Question:2

If the letters of the word RACHIT are arranged in all possible ways as listed in the dictionary. Then what is the rank of the word RACHIT?
[Hint: In each case number of words beginning with A, C, H, and I is 5!]

Answer:

The alphabetical order of the letters of the word RACHIT is A, C, H, I, R, T.
Number of words starting with A=$5!$
Number of words starting with C=$5!$
Number of words starting with H=$5!$
Number of words starting with I=$5!$
For R, the first word is RACHIT
So, the rank of words= $4\ast 5!+1=481$

Question:3

A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

Answer:

Given that candidates can not attempt more than 5 questions from either group and can only attempt
a minimum of 2 questions from either group.
A possible number of questions attempted by each group is:
a) 5 from Group I and 2 from Group II
b) 4 from Group I and 3 from Group II
c) 3 from Group I and 4 from Group II
d) 2 from Group I and 5 from Group II
Total ways= $2\left[{}^6{C}_5{\ast }^6{C}_2{+}^6{C}_4{\ast }^6{C}_3\right]=2\ast 390=780$

Question:4

Out of 18 points in a plane, no three are in the same line except five collinear points. Find the number of lines that can be formed joining the point.

[Hint: Number of straight lines ${=}^{18}{C}_2{-}^5{C}_2+1$ ]

Answer:

Number of straight lines formed by joining 18 points, taking 2 at a time $18C_2$
Number of straight lines formed by joining 5 collinear points, taking 2 at a time = $5C_2$
5 collinear points when joined pairwise give only one line.
Required number of straight lines = $18C_2-5C_2+1=153-10+1=144$

Question:5

We wish to select 6 persons from 8, but if person A is chosen, then B must be chosen. In how many ways can selections be made?

Answer:

Number of persons to be chosen out of 8=6
CASE I: When A is chosen, B must be chosen
Number of ways of selecting 4 more persons from the remaining 6 persons = ${}^{8-2}{C}_{6-2}=6C_4=15$
CASE II: When A is not chosen
Number of ways of selecting 6 persons from the remaining 7 persons = $7C_6=7$

Total ways=$15+7=22$

Question:6

How many committees of five persons with a chairperson can be selected from 12 persons? [Hint: Chairman can be selected in 12 ways and remaining in ${}^{11}{C}_4$.]

Answer:

Out of 12 persons, a chairperson can be selected in ${}^{12}{C}_1$=12 ways.
The remaining 4 persons can be selected out of 11 persons in ${}^{11}{C}_4$=330 ways.
Total number of committee = $12\ast 330=3960$

Question:7

How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

Answer:

Each plate contains 2 different letters followed by 3 different digits
Arrangement of 26 letters taken 2 at a time=${}^{26}{P}_2=26\ast 25=650$Arrangement off off10 digits taken 3 at a time= ${}^{10}{P}_3=10\ast 9\ast 8=720$
Total number of license plates=$650\ast 720=468000$

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 from the lot.

Answer:

2 black balls can be selected from 5 black balls in ${}^5{C}_2$ ways
3 red balls can be selected from 6 red balls in ${}^6{C}_3$ ways
Total number of ways= ${}^5{C}_2{\ast }^6{C}_3=10\ast 20=200$

Question:9

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Answer:

Combination of n things taken r at a time in which 3 things always occur together=${}^{n-3}{C}_{r-3}$
3 things taken together are considered 1 group. Several arrangements of 3 things=$3!$
Now, number of arrangement of r-2 objects=$(r-2)!$
Total number of arrangements=${}^{n-3}{C}_{r-3}\ast (r-2)!\ast 3!$

Question:10

Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.

Answer:

Consonants in TRIANGLE are T, R, N, G, and L.. Vowels are I, A, E
Five consonants can be arranged in 5! ways XCXCXCXCXCX
Arrangements of consonants as C above create gaps marked as X
Now, 3 vowels can be arranged in any three of these 6 gaps in ${}^6{P}_3$ ways
So, to the total number of arrangements=$5!{\ast }^6{P}_3=120\ast 120=14400$

Question:11

Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.

Answer:

A number is divisible by 5 iff at the unit place of the number r, there is 0 or 5. So the unit digit can be filled in 2 ways. Since we have to form 4-digit numbers greater than 6000 and less than 7000. ‘6’ can fill the thousandth place only.

The hundredth and tenth place can be filled together in 8*7 =56 ways. So, the total number of ways = 56*2 =112

Question:12

There are 10 persons named $P_1,P_2,P_3,...P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement, P₁ must occur, whereas P₄ and P₅ do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangements $=7C_4\times 5!$]

Answer:

Given that $P_1,P_2,P_3,...P_{10}$
persons out of which 5 are to be arranged, but $P_1$must occur
whereas P4 and P5 never occur. So, now we only have to select 4 out of 7 people
Number of selections=${}^7{C}_4=35$
Number of the arrangement of 5 persons=$35\ast 5!=35\ast 120=4200$

Question:13

There are 10 lamps in the hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. [Hint: Required number = 2¹⁰ – 1].

Answer:

The hall can be illuminated if at least one lamp is switched on out of 10 lamps Total number of ways
${=}^{10}{C}_1{+}^{10}{C}_2+\cdots \cdot \dots \dots {+}^{10}{C}_{10}=2^{10}-1=1024-1=1023$

Question:14

A box contains two white, three black, and four red balls. In how many ways can three balls be drawn from the box lift least one black ball is to be included in the dradrawwHint: Required number of ways$={}^3{\mathrm{C}}_1\times {}^6{\mathrm{C}}_2+{}^3{\mathrm{C}}_2{\mathrm{x}}^6{\mathrm{C}}_1+{}^3{\mathrm{C}}_3$]

Answer:

Out of 2 white, 3 black, and 4 red balls, we have to draw 3 balls such that at least one black ball is included.
The possibilities are:
a.1 black ball and 2 others b.2 black balls and 1 other (c)3 black balls
Number of selections = $3_{C_1}\ast 6_{C_2}+3_{C_2}\ast 6_{C_1}+3_{C_3}\ast 6_{C_0}$
$=3\ast 15+3\ast 6+1=45+18+1$=64

Question:15

If ${}^n{C}_{r-1}=36$, ${}^n{C}_r=84$and ${}^n{C}_{r+1}=126$, then find ${}^r{C}_2$.
[Hint: From equation $\frac{{}^n{C}_r}{{}^n{C}_{r+1}}\text{ and }\frac{{}^n{C}_r}{{}^n{C}_{r-1}}$ using to find the value of r.]

Answer:

$\begin{array}{c}\frac{n_{C_r}}{n_{C_{r-1}}}=\frac{n-r+1}{r}=\frac{84}{36}=\frac{7}{3}\\ 3n-3r+3=7r\\ 10r-3n=3\dots \dots \dots (i)\\ \frac{n_{C_{r+1}}}{n_{C_r}}=\frac{n-(r+1)+1}{r+1}=\frac{126}{84}\\ \frac{n-r}{r+1}=\frac{3}{2}\\ 2n-2r=3r+3\\ 2n-5r=3\dots \dots .(ii)\end{array}$
Equation (i) + 2 Equation(ii)
10r - 3n +4n - 10r = 3 + 6
n = 9
r =3
${}^r{C}_2{=}^3{C}_2=3$

Question:16

Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8, and 9 where no digits are repeated. [Hint: Besides 4-digit integers greater than 7000, five-digit integers are always greater than 7000.]

Answer:

Since repetitions are not allowed, a maximum of 5- 5-digit numbers can be formed with digits 3, 5,7, 8, and 9.

Since 5-digit numbers are greater than 7000, the number of 5-digit integers = $5\ast 4\ast 3\ast 2\ast 1=120$

A 4-digit number is greater than 7000 if the thousandth place has any of the 7,8, and 9. The remaining 3 places can be filled from the remaining 4 digits in ${}^4{P}_3$ ways

So, the total number of 4-digit integers = $3{\ast }^4{P}_3=3\ast 4\ast 3\ast 2=72$

Total number of integers=$120+72=192$

Question:17

If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

Answer:

If no two lines are parallel, all lines intersect, and no three lines are concurrent.

Two straight lines create one point of intersection
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time
${=}^{20}{C}_2=\frac{20!}{2!18!}=20\ast \frac{19}{2\ast 1}=190$

Question:18

In a certain city, all telephone numbers have six digits, the first two digits always being 41, 42, 46, 62, or 64. How many telephone numbers have all six digits distinct?

Answer:

If the first two digits are 1, the remaining 4 digits can be arranged in ${}^8{P}_4$ ways = 8*7*6*5 = $1680$ ways.

In addition, the first two digits are 42, 46, 62, or 64; the remaining 4 can be arranged in ${}^8{P}_4$ ways = 8*7*6*5 = $1680$ ways.

So, the total number of telephone numbers having all 6 digits different = $5\ast 1680=8400$

Question:19

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 ar,e however compulsory. Determine the number of ways in which the student can make the choice.

Answer:

Since questions are compulsory, the other 2 can be selected from the remaining 3 questions in ${}^3{C}_2$ ways = 3 ways.

Question:20

A convex polygon has 44 diagonals. Find the number of its sides. [Hint: A Polygon of n sides has (${}^n{C}_2-n$a) several diagonals.]

Answer:

Let the convex polygon have n sides.
Number of diagonals = number of ways of selecting two vertices- number of sides

$\begin{array}{c}{n}_{c_2}-n=44\\ \frac{n!}{2!(n-2)!}-n=44\\ \frac{n(n-1)}{2}-n=44\\ n(n-1)-2n=88\\ n^2-3n-88=0\\ (n-11)(n+8)\\ n=11\end{array}$

Question:21

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Answer:

Since 18 mice were placed equally in three groups. Each group had 6 mice
Number of ways of forming 3 groups each with 6 mice= ${}^{18}{C}_6{\ast }^{12}{C}_6{\ast }^6{C}_6=\frac{18!}{6!6!6!}$
But this includes the order in which the three groups are formed, i.e. 3!, which should not be
counted.
So, actual number of ways=$\frac{18!}{6!6!6!3!}$

Question:22

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if
(a) They can be of any c
(b) two must be white and two red and
(c) They must all be of the same colour.

Answer:

Total number of marbles=$6W+5R=11$

a.If they can be of any colour, we have to select 4 out of 11 marbles = $11C_4$
b.2 white marbles can be selected in ${}^6{C}_2$ways and 2 red marbles in ${}^5{C}_2$.
Total number of ways = ${}^6{C}_2{\ast }^5{C}_2=15\ast 10=150$
c. If they all are of the same colour, 4 white out of 6 marbles can be selected in ${}^6{C}_4$ ways and
4 red out of 5 can be selected in ${}^5{C}_4$ ways
Required number of ways=${}^6{C}_4{+}^5{C}_4=15+5=20$

Question:23

In how many ways can a football team of 11 players be selected from 16 players? How many of them will
(i) Include 2 particular players?
(ii) exclude 2 particular players?

Answer:

Number of ways to select a team of 11 players out of 16= ${}^{16}{C}_{11}$
a. If two particular players are included, then 9 more can be selected from the remaining 14 players ${}^{14}{C}_9$
b. If two particular players are excluded, then all 11 players can be selected from the remaining 14
players in ${}^{14}{C}_{11}$

Question:24

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class X II. If there are 20 students in each of these classes, in how many ways can the team be constituted?

Answer:

We have to select at least 5 students from each class out of 20 students from each class
can either select 5 students from class XI and 6 from class XII, or vice versa
Total number of ways= $20c_5\ast 20c_6+20c_6\ast 20c_5=2(20c_6\cdot {20}_{c_5})$

Question:25

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls.

Answer:

i.Team having no girls= ${}^7{C}_5=\frac{7\ast 6}{2!}=21$
ii. Team having at least one boy and one girl= ${}^7{C}_1{\ast }^4{C}_4{+}^7{C}_2{\ast }^4{C}_3{+}^7{C}_3{\ast }^4{C}_2{+}^7{C}_4{\ast }^4{C}_1$
$=7\ast 1+21\ast 4+35\ast 6+35\ast 4=441$
ii I. Team having at least 3 girls = ${}^4{C}_3{\ast }^7{C}_2{+}^4{C}_4{\ast }^7{C}_1=4\ast 21+7=91$

Question:26

If ${}^n{C}_{12}{=}^n{C}_8$, then n is equal to
A. 20
B. 12
C. 6
D. 30

Answer:

The answer is the option (A).
We have,${}^n{C}_{12}{=}^n{C}_8$
$$\begin{gathered} { \\ }^n{C}_{n-12}{=}^n{c}_8 \\ n-12=8 \\ n=20 \end{gathered}$$

Question:27

The number of possible outcomes when a coin is tossed 6 times is
A. 36
B. 64
C. 12
D. 32

Answer:

The answer is the option (b) number of outcomes when a coin is tossed = 2 (head or tail)

Total possible outcomes when a coin is tossed 6 times = $2\ast 2\ast 2\ast 2\ast 2\ast 2=64$

Question:28

The number of different four-digit numbers that can be formed with the digits 2, 3, 4, and 7 and using each digit only once is
A. 120
B. 96
C. 24
D. 100

Answer:

The answer is the option (c)
We have to form 4-digit numbers using 2,3,4, and 7.
Required number of ways = ${}^4{P}_4=4!=24$

Question:29

The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is
A. 432
B. 108
C. 36
D. 18

Answer:

The answer is the option (b)
If we fix 3 at the unit place, the remaining places can be filled in 3! Ways.
Thus ‘3’ appears in unit place in 3! Times.
Similarly, for each of the digits 4,5, and 6. So, sum of digits in unit place = $3!(3+4+5+6)=18\ast 6=108$

Question:30

The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
A. 60
B. 120
C. 7200
D. 720

Answer:

The answer is the option (c)

With 4 vowels and 5 consonants, we have to form words with 2 vowels and 3 consonants.

Total number of ways = ${}^4{C}_2{\ast }^5{C}_3\ast 5!=6\ast 10\ast 120=7200$

Question:31

A five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is
A. 216
B. 600
C. 240
D. 3125
[Hint: 5-digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, and 5 since the sum of digits in these cases is divisible by 3.]

Answer:

The answer is option (a).
A number is divisible by 3 if the sum of its digits is divisible by 3. So, to form several 5-digit which is divisible by 3, we can remove either 0 or 3. If digits 1,2, 3, 4, and 5 are used, then the number of required numbers = 5!
If digits 0,1,2,4,5 are used, then the first place from left can be filled in 4 ways, and the remaining 4 places can be filled in 4! Ways. So, 4*4! Ways.
Total numbers = 120+96 =216

Question:32

Everybody in the room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is
A. 11
B. 12
C. 13
D. 14

Answer:

The answer is the option.

$\begin{array}{rl}& \text{ Number of handshakes }=n_{C_2}=\frac{n!}{(n-2)!2!}=\frac{n(n-1)}{2}=66\\ & n(n-1)=132\\ & n^2-n=132=0\\ & (n-12)(n+11)=0\\ & n=12\end{array}$

Question:33

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line, is
A. 105
B. 15
C. 175
D. 185

Answer:

The answer is the option (d).
Number of ways of selecting 3 points from the given 12 points=${}^{12}{C}_3$
But 3 points from the given 7 collinear points do not form a triangle.
Number of ways of selecting 3 points from seven collinear points = ${}^7{C}_3$
Total number of triangles=${}^{12}{C}_3{-}^7{C}_3=220-35=185$

Question:34

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is
A. 6
B. 18
C. 12
D. 9

Answer:

The answer is the option (b).

We need to select a pair of lines from a set of 4 lines and another pair from another set of 3 lines to form a parallelogram.

Required number = ${}^4{C}_2{\ast }^3{C}_2=6\ast 3=18$

Question:35

The number of ways in which a team of eleven players can be selected from 22 players, always including 2 of them and excluding 4 of the, is
A.${}^{16}{C}_{11}$
B.${}^{16}{C}_5$
C.${}^{16}{C}_9$
D.${}^{20}{C}_9$

Answer:

The answer is the option (c).
We have to select 11 players out of 22.
After excluding the first 4 of them, 18 are available. Amongst these 18, 2 particular players are always included. Thus, we have to select 9 more players from the remaining 16 players in ${}^{16}{C}_9$ ways.

Question:36

The number of 5-digit telephone numbers having at least one of their digits repeated is
A. 90,000
B. 10,000
C. 30,240
D. 69,760

Answer:

The answer is the option (d).
Total number of telephone numbers when there is no restriction = ${10}^5$
Number of telephone numbers with all digits different = ${}^{10}{P}_5$
Required number of ways=${10}^5{-}^{10}{P}_5=100000-10\ast 9\ast 8\ast 7\ast 6$
$=100000-30240=69760$

Question:37

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is
A. 94
B. 126
C. 128
D. None

Answer:

The answer is the option (a).

To satisfy the condition mentioned in the question, we can either select 2 men and 4 women or 3 men and 6 women =

${}^4{C}_2{\ast }^6{C}_4{+}^4{C}_3{\ast }^6{C}_6=6\ast 15+4\ast 1=94$

Question:38

The total number of 9-digit numbers which have all different digits is
A. 10!
B. 9!
C. 9 × 9!
D. 10×10!

Answer:

The answer is option (c).
We have to find a 9-digit number which has all different digits.
The first digit from the left can be filled in 9 ways, excluding 0. After this, nine digits are left, including ‘0’. The remaining 8 places can be filled with these nine digits in ${}^9{P}_8$ ways. So, total numbers= $9{\ast }^9{P}_8=9\ast 9!$

Question:39

The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy an even place, is
A. 1440
B. 144
C. 7!
D.${}^4{C}_4{\times }^3{C}_3$

Answer:

The answer is option (b)
Vowels in the TICLE are A, I, and E, and consonants are R, C, and L.. Vowels occupy 3 even places in 3! Ways and in the remaining 4 places, 4 consonants can be arranged in 4! Ways
Total number of words = $3!\ast 4!=6\ast 24=144$

Question:40

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen, taking at least one green and one blue dye, is
A. 3600
B. 3720
C. 3800
D. 3600
[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are $2^5,2^4$ and $2^3$ respectively.]

Answer:

The answer is option (b).
At least one green dye can be chosen in ${}^5{C}_1{+}^5{C}_2{+}^5{C}_3{+}^5{C}_4{+}^5{C}_5=2^5-1$ ways
At least one blue dye can be chosen in ${}^4{C}_1{+}^4{C}_2{+}^4{C}_3{+}^4{C}_4=2^4-1$ ways
Any number of red dyes can be chosen in ${}^3{C}_0{+}^3{C}_1{+}^3{C}_2{+}^3{C}_3=2^3$ ways
So, total number of selection=$(2^5-1)\ast (2^4-1)\ast \left(2^3\right)=3720$

Question:41

Fill in the Blanks If ${}^n{P}_r=840{,}^n{C}_r=35$, then r = ______.

Answer:

$$\begin{gathered} { \\ }^n{P}_r=840{\text{ and }}^n{C}_r=35{ \\ }^n{P}_r{=}^n{C}_r\ast r! \\ 840=35\ast r! \\ r!=24,r=4 \end{gathered}$$

Question:42

Fill in the Blanks ${}^{15}{C}_8{+}^{15}{C}_9{-}^{15}{C}_6{-}^{15}{C}_7=$______.

Answer:

$=15c_8+15c_9-15c_{15-6}-15c_{15-7}=15c_8+15c_9-15c_8-15c_9=0$

Question:43

Fill in the Blanks
The number of permutations of n different objects, taken r at a time, when repetitions are allowed, is ______.
Answer:

$n\ast n\ast n...rtimes=n^r$

Question:44

Fill in the Blanks The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______.
[Hint: The number of ways of arranging 6 consonants of which two are alike is $\frac{6!}{2!}$ and the number of ways of arranging vowels ${=}^7{P}_6\times \frac{1}{3!}\times \frac{1}{2!}$

Answer:

. Vowels in INTERMEDIATE are I, E, E, I, A, E and consonants N, T, R, M, D, T..
First six consonants are arranged in $\frac{6!}{2!}$ways.
In the 7 gaps, six vowels are arranged in ${}^7{C}_6\ast \frac{6!}{2!3!}$ ways
Total number of words $\frac{6!}{2!}{\ast }^7{C}_6\ast \frac{6!}{2!3!}=360\ast 7\ast 60=151200$

Question:45

Fill in the Blanks
Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.

Answer:

Possible selection can be either 2 red and 1 other or '3 red'.
Required number of ways ${}^5{C}_2{\ast }^7{C}_1{+}^5{C}_3=10\ast 7+10=80$

Question:46

Fill in the Blanks
The number of six-digit numbers, all digits of which are odd, is ______.

Answer:

Odd digits are 1, 3, 5, 7, 9

The required number of numbers $=5\ast 5\ast 5\ast 5\ast 5\ast 5=5^6$

Question:47

Fill in the Blanks. In a football championship, 153 matches were played. Every two teams played one match against each other. The number of teams participating in the championship is ______.

Answer:

Let the number of teams participating in the championship be n.

It is given that every 2 teams played one match against each other.

Total match=${}^n{C}_2=153$

$\frac{n(n-1)}{2}=153$

$n^2-n-306=0$
$$\begin{gathered} (n-18)(n+17)=0 \\ n=18 \end{gathered}$$

Question:48

Fill in the Blanks The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______

Answer:

The number of ways of arranging six '+ 'signs is 1.
In the 7 gaps created, 4 are to be chosen for '-' signs in ${}^7{C}_4$ ways.
These signs can only be arranged in 1 way as they are identical.
Total number of ways $=1{\ast }^7{C}_4\ast 1=35$ ways

Question:49

Fill in the Blanks A committee of 6 is to be chosen from 10 men and 7 women to contain at least 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee?
[Hint: At least 3 men and 2 women: The number of ways = ${}^{10}{C}_3{\times }^7{C}_3{+}^{10}{C}_4{\times }^7{C}_2$.
For 2 particular women to be always there: the number of ways = ${}^{10}{C}_4{+}^{10}{C}_3{\times }^5{C}_1$.
The total number of committees when two particular women are never together = Total – together.]

Answer:

Number of ways of forming a committee of 6 persons containing at least 2 women

${=}^{10}{C}_3{\ast }^7{C}_3{+}^{10}{C}_4{\ast }^7{C}_2=120\ast 35+210\ast 21=8610$

The total number of committees is seen when two particular women are never together

=Total number of ways- Number of ways when 2 particular women are together

= $\left({}^{10}{C}_3{\ast }^7{C}_3{+}^{10}{C}_4{\ast }^7{C}_2\right)-\left({}^{10}{C}_4{+}^{10}{C}_3{\ast }^5{C}_1\right)$

= $4200+4410-(210+600)=7800$

Question:50

Fill in the Blanks A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three barrels are drawn from the box, best one black ball to be included in a drawer, is ______.
Answer:

Required number of ways such that 3 balls are drawn, black balls included
In the draw= ${}^3{C}_1{\ast }^6{C}_2{+}^3{C}_2{\ast }^6{C}_1{+}^3{C}_3$
=$3\ast 15+3\ast 6+1=45+18+1=64$

Question:51

State whether the statements in True or False. There are 12 points in a plane, of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ${}^{12}{C}_2{-}^5{C}_2$.

Answer:

False. Required number of lines= ${}^{12}{C}_2{-}^5{C}_2+1$

Question:52

State whether the statements in True or False.
Three letters can be posted in five letterboxes in $3^5$ ways.

Answer:

False.

Total number of ways of posting 3 letters = $5\ast 5\ast 5=125$

Question:53

State whether the statements in True or False.
In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is ${}^{n-m}{P}_{r-m}{\times }^r{P}_m$.

Answer:

False. In the arrangement of n things taken r at a time in which m occur together, first we
select $(r-m)$ objects from $(n-m)$ objects in ${}^{n-m}{C}_{r-m}$ ways.
If we consider these m things as 1 group number of objects excluding these m objects = $(r-m)$
Now, we have to arrange$(r-m+1)$ objects
Number of arrangements=$(r-m+1)!$
Also, m objects which are considered as 1 group can be arranged in m! ways
Required number of arrangements= ${}^{n-m}{C}_{r-m}\ast (r-m+1!)\ast m!$

Question:54

State whether the statements in True or False.
In a stream, er there are stalls for 12 animals, and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in $3^{12}$ ways.

Answer:

True. In each step, any one of the three animals can be shipped
So, total number of ways of loading$=3\ast 3\ast 3\ast ...12times=3^{12}$

Question:55

State whether the statements in True or False.
If some or all of n objects are taken at a time, the number of combinations is $2^n-1$.

Answer:

True.
If some or all objects are taken at a time, then the number of combinations would be
$n_{C_2}+n_{C_2}+\cdots \dots +n_{C_n}=2^n-1$

Question:56

State whether the statements in True or False.
There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.

Answer:

False.

Several ways of selecting any number of objects from n given identical objects are 1. Now, selecting 0 or more red balls from 4 identical red balls $1+1+1+1+1=5$

Selecting at least 1 black ball from 5 identical black balls =$1+1+1+1+1=5$

So, total ways=$5\ast 5=25$

Question:57

State whether the statements in True or False.
Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side of the table. The number of ways in which the seating arrangements can be made is $\frac{11!}{5!6!}(9!)(9!)$
[Hint: After sending 4 on one side and 3 on the other side, we have to select out of 11; 5 on one side and 6 on the other. Now there are 9 on each side of the long table and each can be arranged in 9! ways.]

Answer:

True
Let the two sides of the table be A and B with 9 seats each.
Let on side A 4 particular guests and on side B 3 particular guests be seated.
Now, for side A, 5 more guests can be selected from the remaining 11 guests in ${}^{11}{C}_5$
ways.
Also, on each side, nine guests can be arranged in 9! ways.
So, total number of ways of arrangement= $11C_5\ast 9!\ast 9!=$\frac{11 !}{5 ! 6 !}(9 !)(9 !)$

Question:58

State whether the statements in True or False. A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.

Answer:

False.
A candidate can attempt questions in either of the following ways :
a.2 from group A and 5 from group B
b.3 from group A and 4 from group B
c.4 from group A and 3 from group B
d.5 from group A and 2 from group B
Number of ways of attempting 7 questions

$\begin{array}{rl}& =6_{C_2}\ast 6_{C_5}+6_{C_3}\ast 6_{C_4}+6_{C_4}\ast 6_{C_3}+6_{C_5}\ast 6_{C_2}\\ & =2(6c_2\ast 6c_5+6c_3\ast 6_{c_4})=2(15\ast 6+20\ast 15)=2(90+3)\end{array}$

Question:59

State whether the statements in True or False.
To fill 12 vacancies, there are 25 candidates, of which 5 are from the scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ${}^5{C}_3{\times }^{22}{C}_9$.

Answer:

True. We can select 3 scheduled candidates out of ${}^5{C}_3$ ways.
And we can select 9 other candidates out of 22 in ${}^{22}{C}_9$ ways.
Total number of selections=${}^5{C}_3{\ast }^{22}{C}_9$

Question:60

There are 3 books on Mathematics, 4 on Physics and 5 on English. How many different collections can be made such that each collection consists of:
$\begin{array}{|llll|}\hline {\text{ C }}_{\mathbf{1}}& & {\mathbf{C}}_2& \\ \text{ (a) }& \begin{array}{c}\text{ One book of each }\\ \text{ subject: }\end{array}& \text{ (i) }& 3968\\ \text{ (b) }& \begin{array}{c}\text{ At least one book }\\ \text{ of each subject: }\end{array}& \text{ (ii) }& 60\\ \text{ (c) }& \begin{array}{c}\text{ At least one book }\\ \text{ of English: }\end{array}& \text{ (iii) }& 3255\\ \hline\end{array}$

Answer:

Number of books on mathematics=3
Number of books of physics = 4
Number of books in English = 5
a.One book of each subject= ${}^3{C}_1{\ast }^4{C}_1{\ast }^5{C}_1=60$
b.At least one book of each subject=$(2^3-1)(2^3-1)(2^5-1)=3225$
c. Adding one book of English=$(2^5-1){.2}^7=3968$
So, a-(ii), b-(iii), c-(i)

Question:61

Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following conditions:
$\begin{array}{|llll|}\hline {\mathbf{C}}_{\mathbf{1}}& & {\mathbf{C}}_2& \\ \text{ (a) }& \begin{array}{c}\text{ Boys and girls }\\ \text{ alternate: }\end{array}& \text{ (i) }& \begin{array}{c}\text{ 5! }\times \\ \text{ 6! }\end{array}\\ \text{ (b) }& \begin{array}{c}\text{ No two girls sit }\\ \text{ together : }\end{array}& \text{ (ii) }& \begin{array}{c}10!\\ -5\\ !6!\end{array}\\ \text{ (c) }& \begin{array}{c}\text{ All the girls sit }\\ \text{ together }\end{array}& \text{ (iii) }& \begin{array}{c}(5!{)}^2\\ +\\ (5!{)}^2\end{array}\\ \text{ (d) }& \begin{array}{c}\text{ All the girls are }\\ \text{ never together : }\end{array}& \text{ (iv) }& \begin{array}{c}2!\\ 5!\\ 5!\end{array}\\ \hline\end{array}$

Answer:

Given that the number of boys=5 anthe d the number of girls
a.Boys and girls alternate=$(5!.5!)+(5!.5!)={(5!)}^2+{(5!)}^2$
b. No two girls sit together=$5!.6!$
c.All the girls sitogether =$!!.5!.5!$
d All the girls are never together :
Total number of boys and girls=$5+5=10$
Number of ways=$10!-5!6!$

Question:62

There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturers is to be formed. Find:

Answer:

Given that the number of professors is 10 and the number of lecturers=20
a.In how many ways committee can be formed= ${}^{10}{C}_2{\times }^{20}{C}_3$
b.In how many ways a particular professor is included= ${}^9{C}_1{\times }^{20}{C}_3$
c.In how many ways a particular lecturer is included= ${}^{10}{C}_2{\times }^{19}{C}_2$
d.In how many ways a particular lecturer is excluded= ${}^{10}{C}_2{\times }^{19}{C}_3$
So a-iv, b-iii, c-ii, d-i

Question:63

Using the digits 1, 2, 3, 4, 5, and 6, 7 seven different digits are from ending
$\begin{array}{|llll|}\hline {\mathbf{C}}_{\mathbf{1}}& & {\mathbf{C}}_{\mathbf{2}}& \\ \text{ (a) }& \begin{array}{c}\text{ how many }\\ \text{ numbers are }\\ \text{ formed? }\end{array}& \text{ (i) }& 840\\ \text{ (b) }& \begin{array}{c}\text{ how many }\\ \text{ numbers are }\\ \text{ exactly divisible }\\ \text{ by 2? }\end{array}& \text{ (ii) }& 200\\ \text{ (c) }& \begin{array}{c}\text{ how many }\\ \text{ numbers are }\\ \text{ exactly divisible }\\ \text{ by 25? }\end{array}& \text{ (iii) }& 360\\ \text{ (d) }& \begin{array}{c}\text{ how many of }\\ \text{ these are exactly }\\ \text{ divisible by 4? }\end{array}& \text{ (iv }& 40\\ \hline\end{array}$

Answer:

Given the number of digits to be formed,
a. How many numbers are formed=${}^7{P}_4=840$
b. How many numbers are exactly divisible by 2?
Number ending with 2= ${}^6{P}_3=120$
Number ending with 4 = ${}^6{P}_3=120$
Numbers ending with 6 = ${}^6{P}_3=120$
So, total numbers=$120+120+120=360$
c. Numbers exactly divisible by 25=40
d. The numbers which have the first 2 digits divisible by 4 are 12, 1, 24, 32, 36,52, 56, 64, 72, 76.So = ${}^{11}{P}_2{\ast }^5{P}_2=200$

Question:64

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
$\begin{array}{|llll|}\hline {\mathbf{C}}_{\mathbf{1}}& & {\mathbf{C}}_2& \\ \text{ (a) }& \begin{array}{c}\text{ 4 letters are used }\\ \text{ at a time }\end{array}& \text{ (i) }& 720\\ \text{ (b) }& \begin{array}{c}\text{ All letters are }\\ \text{ used at a time }\end{array}& \text{ (ii) }& 240\\ \text{ (c) }& \begin{array}{c}\text{ All letters are }\\ \text{ used but the first }\\ \text{ is a vowel }\end{array}& \text{ (iii) }& 360\\ \hline\end{array}$

Answer:

Given that the total number of letters of MONDAY=6
Total vowel=2
a. When 4 letters are used at a time = ${}^6{P}_4=360$
b. All the letters are used at a time = ${}^6{P}_6=720$
c. All letters are used, but the first is a vowel=$2.5!$