# XAT Permutation and Combination Questions [Download PDF]

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**Question 1: **Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?

a) 4320

b) 720

c) 2160

d) 120

e) None of these

**1) Answer (A)**

**Solution:**

Word – DIRECTOR

So “I,E,O” are there are 3! ways to arrange the vowels

Now “D,R,C,T,R” are the remaining alphabets ,

Condition is that the vowels should always be together so we can assume the vowels as a single alphabet/unit say “X” (‘X’=’I,E,O’) so now we have a new word – “D,R,C,T,R,X”

Possible arrangements for this word = 6!

Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of ‘DRCTRX’) $\times$ (Possible arrangements of vowels)

= 6! $\times$ 3! = $720 \times 6 = 4320$

=> Ans – (A)

**Question 2: **In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?

a) 1/55

b) 54/55

c) 45/55

d) 3/55

e) None of these

**2) Answer (B)**

**Solution:**

Total number of oranges in the box = 12

Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$

= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$

Number of oranges which became bad = $\frac{12}{3}=4$

Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$

Number of desired selection of oranges, n(E) = 220 – 4 = 216

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

= $\frac{216}{220}= \frac{54}{55}$

=> Ans – (B)

**Instructions**

Study the information carefully to answer the following questions:

A bucket contains 8 red, 3 blue and 5 green marbles.

**Question 3: **If 3 marbles are drawn at random, what is the probability that none is red ?

a) ${3 \over 8}$

b) ${1 \over {16}}$

c) ${1 \over {10}}$

d) ${3 \over {16}}$

e) None of these

**3) Answer (C)**

**Solution:**

Number of ways of drawing 3 marbles out of 16

$n(S) = C^{16}_3 = \frac{16 \times 15 \times 14}{1 \times 2 \times 3}$

= $560$

Out of the three drawn marbles, none is red, i.e., they will be either blue or green.

=> $n(E) = C^8_3 = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}$

= $56$

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

= $\frac{56}{560} = \frac{1}{10}$

**Question 4: **If 2 marbles are drawn at random, what is the probability that both are green?

a) ${1 \over 8}$

b) ${5 \over {16}}$

c) ${2 \over 7}$

d) ${3 \over 8}$

e) None of these

**4) Answer (E)**

**Solution:**

Number of ways of drawing 2 marbles out of 16

$n(S) = C^{16}_2 = \frac{16 \times 15}{1 \times 2}$

= $120$

Out of the two drawn marbles, both are green

=> $n(E) = C^5_2 = \frac{5 \times 4}{1 \times 2}$

= $10$

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

= $\frac{10}{120} = \frac{1}{12}$

**Question 5: **If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?

a) ${{11} \over {16}}$

b) ${3 \over {16}}$

c) ${11 \over {72}}$

d) ${3 \over {65}}$

e) None of these

**5) Answer (D)**

**Solution:**

Number of ways of drawing 4 marbles out of 16

=> $n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$

= $1820$

Out of the four drawn marbles, 2 are red and 2 are blue.

=> $n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$

= $28 \times 3 = 84$

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

= $\frac{84}{1820} = \frac{3}{65}$

**Instructions**

From the following, differen committees are to be made as per the requirement given in each question.

In how many different ways can it de done? 10 men and 8 women out of which 5 men are teachers, 3 men doctors and businessmen. Among the women, 3 are 2 are teachers, 2 doctors, 2 researchers and 1 social worker.

**Question 6: **A Committee of 3 in which there is no teacher and no doctor.

a) 100

b) 120

c) 10

d) 12

e) None of these

**6) Answer (C)**

**Solution:**

Out of 10 men, Teachers = 5

Doctors = 3 and Business man = 2

Out of 8 women, Teachers = 3 and Doctors = 2

Researchers = 2 and Social worker = 1

Now, no teacher and doctor should be selected.

=> Remaining members (both men and women) = (2) + (2 + 1) = 5

Thus, number of ways of selecting 3 people out of 5

= $C^5_3 = \frac{5 \times 4 \times 3}{1 \times 2 \times 3} = 10$

**Question 7: **A Committee of 7.

a) 31824

b) 1200

c) 9600

d) 15912

e) None of these

**7) Answer (A)**

**Solution:**

There are 10 men and 8 women, Total = 18

Number of ways of selecting 7 members out of 18.

= $C^{18}_7 = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}$

= $18 \times 17 \times 4 \times 13 \times 2 = 31824$

**Question 8: **A Committee is 5 in which 2 men teachers, 2 women teachers and 1 doctor are there.

a) 75

b) 150

c) 214

d) 20

e) None of these

**8) Answer (B)**

**Solution:**

Out of 10 men, Teachers = 5

Doctors = 3 and Business man = 2

Out of 8 women, Teachers = 3 and Doctors = 2

Researchers = 2 and Social worker = 1

Number of ways of selecting 2 men teachers, 2 women teachers and 1 doctor

= $C^5_2 \times C^3_2 \times C^5_1$

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

= $10 \times 3 \times 5 = 150$

**Question 9: **A Committee of 4 in which at least 2 women are there.

a) 1260

b) 1820

c) 3060

d) 1890

e) None of these

**9) Answer (D)**

**Solution:**

Out of 10 men, Teachers = 5

Doctors = 3 and Business man = 2

Out of 8 women, Teachers = 3 and Doctors = 2

Researchers = 2 and Social worker = 1

Number of ways of selecting 4 members in which at least 2 are women

= (2 men , 2 women) + (1 men , 3 women) + (0 men , 4 women)

= $(C^{10}_2 \times C^8_2) + (C^{10}_1 \times C^8_3) + (C^{10}_0 \times C^8_4)$

= $(\frac{10 \times 9}{1 \times 2} \times \frac{8 \times 7}{1 \times 2}) + (\frac{10}{1} \times \frac{8 \times 7 \times 6}{1 \times 2 \times 3}) + (\frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4})$

= $(45 \times 28) + (10 \times 56) + (70)$

= $1260 + 560 + 70 = 1890$

**Question 10: **A Committee of 5 in which 3 men and 2 women are there.

a) 3360

b) 8568

c) 4284

d) 1680

e) None of these

**10) Answer (A)**

**Solution:**

Out of 10 men, Teachers = 5

Doctors = 3 and Business man = 2

Out of 8 women, Teachers = 3 and Doctors = 2

Researchers = 2 and Social worker = 1

Number of ways of selecting 3 men and 2 women.

= $C^{10}_3 \times C^8_2$

= $\frac{10 \times 9 \times 8}{1 \times 2 \times 3} \times \frac{8 \times 7}{1 \times 2}$

= $120 \times 28 = 3360$