Permutation and Combination Questions for MAH-CET

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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 1: 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

1) 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 2: A Committee of 7.

a) 31824

b) 1200

c) 9600

d) 15912

e) None of these

2) 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 3: 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

3) 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 4: A Committee of 4 in which at least 2 women are there.

a) 1260

b) 1820

c) 3060

d) 1890

e) None of these

4) 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 5: 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

5) 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.

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

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

= $120 \times 28 = 3360$

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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 6: If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?

a) 4/15

b) 17/280

c) 6/91

d) 11/15

e) None of these

6) Answer (C)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $P(S) = C^{15}_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$

= $1365$

P(E) = Favorable outcomes

= Selecting 1 green, 2 blue and 1 red marble.

=> $P(E) = C^2_1 \times C^6_2 \times C^3_1$

= $2 \times \frac{6 \times 5}{1 \times 2} \times 3$

= $90$

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

= $\frac{90}{1365} = \frac{6}{91}$

Question 7: If two marbles are picked at random, what is the probability that either both are red or both are green ?

a) 3/5

b) 4/105

c) 2/7

d) 5/91

e) None of these

7) Answer (B)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 2 marbles at random out of 15

=> $P(S) = C^{15}_2 = \frac{15 \times 14}{1 \times 2}$

= $105$

P(E) = Favorable outcomes

= Selecting 2 green or 2 red marbles.

=> $P(E) = C^2_2 + C^3_2$

= $1 + \frac{3 \times 2}{1 \times 2}$

= $4$

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

= $\frac{4}{105}$

Question 8: If four marbles are picked at random, what is the probability that at least one is yellow ?

a) 91/123

b) 69/91

c) 125/143

d) 1/3

e) None of these

8) Answer (B)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $P(S) = ^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$

= $1365$

Let no yellow marble is selected.

P(E) = Favorable outcomes

= Selecting 4 out of 11 marbles.

=> $P(E) = ^{11}C_4$

= $\frac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4}$

= $330$

$\therefore$ Required probability = $1 – \frac{P(E)}{P(S)}$

= $1 – \frac{330}{1365} = 1 – \frac{22}{91}$

= $\frac{91 – 22}{91} = \frac{69}{91}$

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

9) Answer (C)

Solution:

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

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

Question 10: If two marbles are picked at random, what is the probability that both are green ?

a) 2/15

b) 1/15

c) 2/7

d) 1

e) None of these

10) Answer (E)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 2 marbles at random out of 15

=> $P(S) = C^{15}_2 = \frac{15 \times 14}{1 \times 2}$

= $105$

P(E) = Favorable outcomes

= Selecting 2 green marbles.

=> $P(E) = C^2_2 = 1$

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

= $\frac{1}{105}$

Instructions

Study the given information carefully and answer the questions that follow:
An urn contains 6 red, 4 blue, 2 green, 3 yellow marbles.

Question 11: If two marbles are picked at random what is the probability that either both are green or both are yellow ?

a) $\frac{5}{91}$

b) $\frac{1}{35}$

c) $\frac{1}{3}$

d) $\frac{4}{105}$

e) None of these

11) Answer (D)

Solution:

Number of ways of selecting 2 green marbles or 2 yellow marbles = $^2C_2+^3C_2$ = 1+3 = 4

Probability = $4/^{15}C_2$  = 4/155

Question 12: If four marbles are picked at random what is the probability that one is green two are blue and one is red ?

a) $\frac{24}{455}$

b) $\frac{13}{35}$

c) $\frac{11}{15}$

d) $\frac{7}{91}$

e) None of these

12) Answer (A)

Solution:

One green marble can be selected in $^2C_1$ = 2 ways.

Two blue marbles can be selected in $^4C_2 = 6$ ways

Six red marbles can be selected in $^6C_1 = 6$ ways.

Probability = $\frac{2*6*6}{^{15}C_4}$ = 24/455

Question 13: If three marbles are picked at random what is the probability that two are blue and one is yellow ?

a) $\frac{3}{91}$

b) $\frac{1}{5}$

c) $\frac{18}{455}$

d) $\frac{7}{15}$

e) None of these

13) Answer (C)

Solution:

Total number of ways to pick up 3 marbles = $^{15}C_3$ = 455

Total number of favourable events = $^4C_2*3$ = 18

Hence, probability = $\frac{18}{455}$

Question 14: If two marbles are picked at random what is the probability that both are red ?

a) $\frac{1}{6}$

b) $\frac{1}{3}$

c) $\frac{2}{15}$

d) $\frac{2}{5}$

e) None of these

14) Answer (E)

Solution:

Required probability would be $^6C_2/^{15}C_2$ = 1/7 as we are selecting 2 balls out of 6 red balls out og the total possibility set of 2 balls out of 15 balls.

Question 15: If four marbles are picked at random what is the probability that at least one is blue ?

a) $\frac{4}{15}$

b) $\frac{69}{91}$

c) $\frac{11}{15}$

d) $\frac{22}{91}$

e) None of these

15) Answer (B)

Solution:

Total number of events = $^{15}C_4$ = 1365
Probability of at least 1 Blue = 1 – (Probability of 0 blue)
Probability of 0 blue = $\frac{^{11}C_4}{1365}$ = $\frac{22}{91}$
Probability of at least 1 Blue = 1 – $\frac{22}{91}$ = $\frac{69}{91}$

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