**Permutation and Combination Questions for XAT Set-2 PDF**

Download important Permutation and Combination Questions for XAT Set-2 PDF based on previously asked questions in the XAT exam. Practice Permutation and Combination Questions Set-2 PDF for the XAT exam.

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**Instructions**Instructions: Study the given information carefully to answer the questions that follows.

An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.

**Question 1: **If two marbles are drawn at random, what is the probability that both are red or at least one is red?

a) $\frac{26}{91}$

b) $\frac{1}{7}$

c) $\frac{199}{364}$

d) $\frac{133}{191}$

e) None of these

**Question 2: **What is the number of words formed from the letters of the word ‘JOKE’ So that the vowels and consonants alternate?

a) 4

b) 8

c) 12

d) 18

e) None of these

**Instructions**

Study the given information carefully and answer the question that follow:

A committee of five members is to be formed out of 3 trainees 4 professors and 6 research associates In how many different ways can this be done if____

**Question 3: **The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors ?

a) 12

b) 13

c) 24

d) 52

e) None of these

**Question 4: **In how many different ways can the letters of the word BAKERY be arranged ?

a) 2400

b) 2005

c) 720

d) 5040

e) None of these

**Instructions**

Answer the following questions based on the information given below.

A bowl contains 4 red, 3 green, 2 blue and 5 black marbles.

**Question 5: **If four marbles are drawn at random, what is the probability that two are red and two are blue ?

a) $\frac{6}{1001}$

b) $\frac{1}{143}$

c) $\frac{1}{13}$

d) $\frac{6}{91}$

e) None of these

**Question 6: **In a sample , if a person is picked up randomly, the probability that the person is a smoker is $\frac{3}{5}$, and that of the person being male is $\frac{1}{2}$ .What is the probability that the person is both male as well as a smoker ?

a) $\frac{10}{11}$

b) $\frac{1}{5}$

c) $\frac{3}{5}$

d) Cannot be determined

e) None of these

**Question 7: **A committee of 6 members is to be selected from a group of 8 men and 6 women in such as way that at least 3 men are there in the committee. In how many different ways can it be done ?

a) 2506

b) 2534

c) 1120

d) 1050

e) None of these

**Question 8: **In how many different ways can the letters of the word ‘QUOTED’ be arranged

a) 720

b) 360

c) 1440

d) 320

e) None of these

**Instructions**

Study the given information carefully and answer the questions that follow:

An urn contains 3 red, 6 blue, 2 green and 4 yellow marbles.

**Question 9: **If three marbles are picked at random, what is the probability that two are blue and one is yellow ?

a) 2/15

b) 6/91

c) 12/91

d) 3/15

e) None of these

**Question 10: **In how many different ways can the letters of the word “PRIDE” be arranged ?

a) 60

b) 120

c) 15

d) 360

e) None of these

XAT Decision making practice questions

**Answers & Solutions:**

**1) Answer (E)**

Out of two red marbles two can be choosen in ^{2}C_{2} ways

Out of two red marbles one can be choosen in ^{2}C_{1} ways

Sum of blue, yellow and green marbles = 12

Out of 12 marbles one can be choosen in ^{12}C_{1} ways

The probability that both are red or atleast one is red = [^{2}C_{2} + (^{2}C_{1} x ^{12}C_{1})]/^{14}C_{2} = $\frac{25}{91}$

**2) Answer (B)**

Word name: ‘JOKE’ Vowels: O, E

Consonants: J, K

∴ Possible arrangement beginning with consonant: JOKE, KOJE, JEKO, KEJO = 4 Numbers

beginning with vowel: OJEK, OKEJ, EJOK, EKOJ = 4 Numbers

Required number = 4+4 = 8 numbers

**3) Answer (A)**

4 professors and 1 resarch associate can be selected in $^4C_4*^6C_1 = 6$ ways

Similarly 3 trainees and 2 professors would be selected in $^3C_3*^4C_2$ ways = 6

Total ways = 6+6 = 12

**4) Answer (C)**

As there are 6 different letters in the word “Bakery” hence, total number of ways of arranging it will be = 6! = 720

**5) Answer (A)**

Probability of drawing two red and two blue balls = $^4C_2 * ^2C_2 / ^{14}C_4$ = 6 / 1001

Option a) is the correct answer.

**6) Answer (D)**

Let’s assume the sample size is 100. Let the number of male smokers be x.

Total number of smokers = 3/5 * 100 = 60

Number of men = 1/2 * 100 = 50

Hence, number of male non-smokers is 50-x. Number of female smokers is 60-x and number of female non-smokers is x-10.

Hence, probability of a person picked at random being a smoker and a male = x/100

As we do not know the value of x, we cannot determine the probability. Hence, option D.

**7) Answer (B)**

There are a total of 8 men and 6 women.

Number of ways in which the committee has exactly 3 men is $^8C_3 \times ^6C_3 = 1120$

Number of ways in which the committee has exactly 4 men is $^8C_4 \times ^6C_2 = 1050$

Number of ways in which the committee has exactly 5 men is $^8C_5 \times ^6C_1 = 336$

Number of ways in which the committee has exactly 6 men is $^8C_6 \times ^6C_0 = 28$

Hence, the total is $1120 + 1050 + 336 + 28 = 2534$

**8) Answer (A)**

Total number of arrangement= 6!=720

**9) Answer (C)**

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 3 marbles at random out of 15

=> $P(S) = ^{15} C_3 = \frac{15 \times 14 \times 13}{1 \times 2 \times 3}$

= $455$

P(E) = Favorable outcomes

= Selecting 2 blue and 1 yellow marble.

=> $P(E) =C^6_2 \times C^4_1$

= $\frac{6 \times 5}{1 \times 2} \times 4$

= $60$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{60}{455} = \frac{12}{91}$

**10) Answer (B)**

The word ‘PRIDE’ consists of 5 distinct letters

=> Number of arrangements = $5!$

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

We hope this Permutation and Combination Questions Set-2 PDF for XAT with Solutions will be helpful to you.