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# Permutation and Combinations Questions for CAT

Permutation and Combinations is a key topic in the CAT Quants section. Make sure you are aware of all the Important Concepts in CAT Permutation and Combinations. You can also check out these CAT Permutation and Combinations questions from the CAT Previous year papers. This post will look into some important Permutation and Combinations questions for CAT. These are good sources of practice for CAT 2022 preparation. If you want to practice these questions, you can download these Important Permutation and Combinations Questions for CAT (with detailed answers) PDF and the video solutions below, which are completely Free.

Question 1:Â Refer to the figure given below. AB, CD and EF are three parallel paths. A person starts from AB to reach EF by moving in four steps – moving from AB to $O_1$ in step 1, from $O_1$ to CD in step 2, from CD to $O_2$ in step 3 and from $O_2$ to EF in step 4. If he takes a curved path in one step, he cannot take a curved path in the next step. In how many ways the person can reach EF from AB ?

a)Â 128

b)Â 256

c)Â 32

d)Â 64

Solution:

The total number of possible ways from AB to EF = $4\times 4\times 4\times 4=256$

If a curved path is taken in a step, then the next step cannot be a curved path.

Let C denotes a curved path and S denotes a straight path.

Thus, the invalid paths are as follows: CCCC, CCCS, CCSC, CSCC, SCCC, CCSS, SCCS, SSCC. Thus, a total of 8 paths.

Total number of arrangements for these 8 paths = $8\times 2\times 2\times 2\times 2=128$

Thus, the total number of valid paths = $256-128=128$

Hence, the answer is option A.

Question 2:Â A small store has five units of a new phone model in stock: two white, two black, and one red. Three customers arrive at the shop to buy a unit each. Each one has a pre- determined choice of the colour and will not buy a unit of any other colour. All the three customers are equally likely to have chosen any of the three colours. What is the probability that the store will be able to satisfy all the three customers?

a)Â $\frac{4}{5}$

b)Â $\frac{7}{9}$

c)Â $\frac{2}{3}$

d)Â $\frac{8}{9}$

e)Â $\frac{1}{3}$

Solution:

Number of white phones = 2

Number of black phones = 2

Number of red phones = 1

customer 1 will have 3 choices

customer 2 will have 3 choices

customer 3 will have 3 choices

Hence total choices = 3 x 3 x 3 = 27

The cases not possible = BBB, RRR,WWW, RRB,RBR,BRR, RRW,RWR, WRR

Possible cases = 18

Probability = 18/27 = 2/3

Question 3:Â In how many different ways, can the letters of the words EXTRA be arranged so that the vowels are never together ?

a)Â 168

b)Â 48

c)Â 120

d)Â 72

Solution:

Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters.

These letters can be arranged in $4! = 24$ ways

The letters EA may be arranged amongst themselves in $2!=2$ ways.

Number of arrangements having vowels together = $(24 \times 2) = 48$ ways

Total arrangements of all lettersÂ = $5!=120$ ways

=> Number of arrangements not having vowels togetherÂ = $(120 – 48)Â = 72$ ways

=> Ans – (D)

Question 4:Â There are six teachers. Out of them, two teach physics, other two teach Chemistry and the rest two teach Mathematics. They have to stand in a row such that Physics, Chemistry and Mathematics teachers are always in a set. The numberof ways in which they can do, is :

a)Â 24

b)Â 48

c)Â 36

d)Â 12

Solution:

There are 2 physics teacher.They can stand in a row in $P(2,2)=2!=2\times1=2$ ways

Similarly, the two Chemistry and Mathematics teacher can stand in 2 ways each.

These three sets can be arranged in themselves in $3!=3\times2\times1=6$ ways

=> Required number of ways = $2\times2\times2\times6=48$

=> Ans – (B)

Question 5:Â Brother and sister both appear for an interview. The probability of the selection of brother is $\frac{1}{8}$ while the probability of rejection of sister is $\frac{4}{5}$ What is the probability that only one of them is selected ?

a)Â $\frac{11}{40}$

b)Â $\frac{7}{40}$

c)Â $\frac{5}{13}$

d)Â $\frac{1}{10}$

Solution:

Probability of brother’s selection = $\frac{1}{8}$ and probability of brother’s rejection = $\frac{7}{8}$

Probability of sister’s selection = $\frac{1}{5}$ and probability of sister’s rejection = $\frac{4}{5}$

Probability that only one of them is selected = (prob. that brother is selected) Ã— (prob. that sister is not selected) + (Prob. that brother is not selected) Ã— (Prob. that sister is selected)

= $(\frac{1}{8})(\frac{4}{5})+(\frac{7}{8})(\frac{1}{5})$

= $\frac{4}{40}+\frac{7}{40}=\frac{11}{40}$

=> Ans – (A)

Question 6:Â In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women ?

a)Â 5041

b)Â 5040

c)Â 11670

d)Â 11760

Solution:

Number of waysÂ a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women

= $C^{10}_6\timesÂ C^8_5$

=Â $C^{10}_4\times C^8_3$

= $\frac{10\times9\times8\times7}{1\times2\times3\times4}\times\frac{8\times7\times6}{1\times2\times3}$

= $(10\times3\times7)\times(8\times7)$

= $210\times56=11760$

=> Ans – (D)

Question 7:Â Let n be the number of ways in which 5 men and 6 women can stand in a queue such that all the women stand consecutively. Let m be the number of ways in which the same 11 persons can stand in a queue such that exactly 5 women stand consecutiv. ely. The value of $\frac{m}{n}$ is

a)Â 5

b)Â 6

c)Â $\frac{5}{6}$

d)Â $\frac{6}{5}$

Solution:

For n, we consider B as a group that has all 6 women. Total arrangement of B and 5 men is 6!. Further within B women can rearrange themselves in 6! ways.

Thus n = 6! $\ \times\$6!

For m, consider a groupÂ $B_1$ which has 5 women andÂ $B_2$ has remaining 1 woman. Arrangement ofÂ $B_1$,$B_2$ and 5 men can be done in 7! ways. From this we need to subtract the permutation in whichÂ $B_1$ andÂ $B_2$ are together. Which is $2!\times\ 6!$Â . Total waysÂ $=\ 7!\ -\ 2!\times 6!$ .Which equalsÂ $5\times\ 6!$. Futhermore in $B_1$ the selection and arrangement of 5 women can be done inÂ $_5^6C\times\ \ 5!$ = 6! ways.

m =Â $5\times\ 6!\times\ 6!$

$\ \frac{\ m}{n}=5$

Question 8:Â How many distinct 5 x 5 matrices are there such that each entry is either 0 or 1 and each row sum and each column sum is 4?

a)Â 64

b)Â 32

c)Â 120

d)Â 96

Solution:

With either 0 or 1 as an entry, a sum of 4 is possible only with four 1s and one Zero.

So, each column and row must have four 1s and one 0.

Let us consider the first row, we can put the O in any of the 5 places.

So, number of ways of placing 1 zero in the first row=5.

In the second row, we cannot put the zero in the same column as that put by the first, as it will result in two )s in the same column. So, the second 0 can be but in any of the remainingÂ 4 places.

Similarly, for the third zero, we will have 3 ways to do so. And so on.

So, total number of ways to put 5 zeroes in the 5$\times\ 5$ matrix is 5*4*3*2*1= 120.

Question 9:Â Ashok has a bag containing 40 cards, numbered with the integers from 1 to 40. No two cards are numbered with the same integer. Likewise, his sister Shilpa has another bag containing only five cards that are numbered with the integers from 1 to 5, with no integer repeating. Their mother, Latha, randomly draws one card each from Ashokâ€™s and Shilpaâ€™s bags and notes down their respective numbers. If Latha divides the number obtained from Ashokâ€™s bag by the number obtained from Shilpaâ€™s, what is the probability that the remainder will not be greater than 2?

a)Â 0.8

b)Â 0.91

c)Â 0.73

d)Â 0.94

e)Â 0.87

Solution:

The number of ways of selecting one card from Ashok’s bag and other from Shilpa bag =Â $40_{C_1}\times\ 5_{C_1}$ = 200
Now, if the card taken from Shilpa’s bag shows 1, then 1 will divide all the numbers on Ashok’s card. Hence, the number of ways = 40
If the card taken from Shilpa’s bag shows 2, then the remainder will be either 0 or 1. Hence, the number of ways = 40
If the card taken from Shilpa’s bag shows 3, then the remainder will be 0, 1 or 2. Hence, the number of ways = 40
If the card taken from Shilpa’s bag shows 4, then the remainder will be 0, 1, 2 or 3. So the numbers having 3 as remainder will be rejected. So the number of form 4n+3 will be rejected. Total number of such numbers =Â $\frac{\left(39-3\right)}{4}+1$ = 10
If the card taken from Shilpa’s bag shows 5, then the remainder will be 0, 1, 2, 3 or 4. So the numbers having 3 or 4Â as remainder will be rejected. So the number of form 5n+3, 5n+4Â will be rejected. Total number of such terms = $\frac{\left(39-3\right)}{4}+1$ = 10
The numbers left = 40-10 = 30
The total numbers having 5n+3 form =Â $\frac{\left(39-4\right)}{5}+1\ =\ 8$
The total numbers having 5n+4 form =Â $\frac{\left(38-3\right)}{5}+1\ =\ 8$
The numbers left = 40-8-8=24

Hence, the probability =Â $\frac{\left(40+40+40+30+24\right)}{200}=\frac{174}{200}=0.87$

Question 10:Â A number lock consists of 3 rings each marked with 10 different numbers. In how many cases the lock cannot be opened?

a)Â $3^{10}$

b)Â $10^{3}$

c)Â 30

d)Â 999

Solution:

The total number of combinations = 10*10*10Â  = 1000Â  Â  (There are 10-digits 0 to 9 on each ring)

Since there is only 1 right combination, the number of cases in which the lock cannot be opened = 1000-1=999

Question 11:Â There are 10 stations on a railway line. The number of different journey tickets that are required by the authorities is

a)Â 10!

b)Â 90

c)Â 81

d)Â 10

Solution:

The number of different journey tickets that are required by the authorities = The number of selecting 2 stationsÂ $\times\$ 2Â  Â (For every pair stations say A and B, there are two tickets required A to B and B to A)

= $\\^{10}C_2$ * 2 = 45*2=90

Question 12:Â Four stacks containing an equal number of chips are to be made from 11 orange, 9 white, 13 black and 7 yellow chips. If all of these chips are used and each stack contains at least one chip of each colour, what is the maximum number of whiteÂ chips in any one stack?

a)Â 3

b)Â 4

c)Â 5

d)Â 6

Solution:

Total number of chips = 11+9+13+7=40

Total number of chips in each stack = 40/4 = 10

Total number of white chips = 9,

The maximum number of white chips in a stack will be obtained when 1 white chip given to each of 3 stacks and 6 white chips to 1 stack.

Question 13:Â A and B throw one dice for a stake of Rs.11, which is to be won by the player who first throws a six.
The game ends when stake is won by A or B. If A has the first throw, what are their respective expectations?

a)Â 5 and 6

b)Â 6 and 5

c)Â 11 and 0

d)Â 10 and 1

Solution:

The probability of throwing a 6 =Â $\ \frac{\ \ 1}{\ 6}$

The probability of not throwing a 6 =Â $\ \frac{\ \ 5}{\ 6}$

The probability that A wins =Â $\ \frac{\ \ 1}{\ 6}+\ \frac{\ 5}{\ 6}\times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 1}{6}+\ \frac{\ 5}{6}\ \times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 5}{6}\times\ \frac{\ 1}{6}\ +\ ……………$

=Â $\ \frac{\ 1}{6}\left(1+\left(\ \frac{\ 5}{6}\right)^2+\left(\ \frac{\ 5}{6}\right)^4………\right)$

=Â $\ \ \frac{\ 1}{6}\left(\ \dfrac{\ 1}{\ 1-\ \frac{\ 25}{36}}\right)$ =Â $\ \frac{\ 6}{11}$

The expected return of AÂ Â = 11*$\ \frac{\ 6}{11}$ = 6

Hence the expected return of B = 11-6=5

Question 14:Â There are 10 stations on a railway line. The number of different journey tickets that are required by the authorities is

a)Â 10!

b)Â 90

c)Â 81

d)Â 10

Solution:

The number of different journey tickets that are required by the authorities = The number of selecting 2 stationsÂ $\times\$ 2Â  Â (For every pair stations say A and B, there are two tickets required A to B and B to A)

= $\\^{10}C_2$ * 2 = 45*2=90

Question 15:Â If Swamy has two children and he truthfully answers yes to the question “Is at least one of your children a girl?” what is the probability that both his children are girls?

a)Â $\frac{1}{2}$

b)Â $\frac{1}{3}$

c)Â 1

d)Â 0

Probability =Â $\frac{\left(Number\ of\ favourable\ cases\right)}{Total\ number\ of\ cases}$