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
=
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.
=
=
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
=
=
=
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)
=
=
=
=
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.
=
=
=
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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(E) = Favorable outcomes
= Selecting 1 green, 2 blue and 1 red marble.
=>
=
=
=
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(E) = Favorable outcomes
= Selecting 2 green or 2 red marbles.
=>
=
=
=
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
=>
=
Let no yellow marble is selected.
P(E) = Favorable outcomes
= Selecting 4 out of 11 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
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(E) = Favorable outcomes
= Selecting 2 blue and 1 yellow marble.
=>
=
=
=
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(E) = Favorable outcomes
= Selecting 2 green marbles.
=>
=
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)
b)
c)
d)
e) None of these
11) Answer (D)
Solution:
Number of ways of selecting 2 green marbles or 2 yellow marbles =
Probability =
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)
b)
c)
d)
e) None of these
12) Answer (A)
Solution:
One green marble can be selected in
Two blue marbles can be selected in
Six red marbles can be selected in
Probability =
Question 13: If three marbles are picked at random what is the probability that two are blue and one is yellow ?
a)
b)
c)
d)
e) None of these
13) Answer (C)
Solution:
Total number of ways to pick up 3 marbles =
Total number of favourable events =
Hence, probability =
Question 14: If two marbles are picked at random what is the probability that both are red ?
a)
b)
c)
d)
e) None of these
14) Answer (E)
Solution:
Required probability would be
Question 15: If four marbles are picked at random what is the probability that at least one is blue ?
a)
b)
c)
d)
e) None of these
15) Answer (B)
Solution:
Total number of events =
Probability of at least 1 Blue = 1 – (Probability of 0 blue)
Probability of 0 blue =
Probability of at least 1 Blue = 1 –