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# Permutation and Combination Questions for MAH-CET

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:Â A Committee of 3 in which there is no teacher and no doctor.

a)Â 100

b)Â 120

c)Â 10

d)Â 12

e)Â None of these

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:Â A Committee of 7.

a)Â 31824

b)Â 1200

c)Â 9600

d)Â 15912

e)Â None of these

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

a)Â 75

b)Â 150

c)Â 214

d)Â 20

e)Â None of these

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 selectingÂ 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:Â A Committee of 4 in which at least 2 women are there.

a)Â 1260

b)Â 1820

c)Â 3060

d)Â 1890

e)Â None of these

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:Â A Committee of 5 in which 3 men and 2 women are there.

a)Â 3360

b)Â 8568

c)Â 4284

d)Â 1680

e)Â None of these

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

a)Â 4/15

b)Â 17/280

c)Â 6/91

d)Â 11/15

e)Â None of these

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

a)Â 3/5

b)Â 4/105

c)Â 2/7

d)Â 5/91

e)Â None of these

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

a)Â 91/123

b)Â 69/91

c)Â 125/143

d)Â 1/3

e)Â None of these

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

a)Â 2/15

b)Â 6/91

c)Â 12/91

d)Â 3/15

e)Â None of these

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

a)Â 2/15

b)Â 1/15

c)Â 2/7

d)Â 1

e)Â None of these

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

a)Â $\frac{5}{91}$

b)Â $\frac{1}{35}$

c)Â $\frac{1}{3}$

d)Â $\frac{4}{105}$

e)Â None of these

Solution:

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

Probability = $4/^{15}C_2$ Â = 4/155

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

a)Â $\frac{24}{455}$

b)Â $\frac{13}{35}$

c)Â $\frac{11}{15}$

d)Â $\frac{7}{91}$

e)Â None of these

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

a)Â $\frac{3}{91}$

b)Â $\frac{1}{5}$

c)Â $\frac{18}{455}$

d)Â $\frac{7}{15}$

e)Â None of these

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

a)Â $\frac{1}{6}$

b)Â $\frac{1}{3}$

c)Â $\frac{2}{15}$

d)Â $\frac{2}{5}$

e)Â None of these

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

a)Â $\frac{4}{15}$

b)Â $\frac{69}{91}$

c)Â $\frac{11}{15}$

d)Â $\frac{22}{91}$

e)Â None of these

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