header

GATE Online Mock Test Series FREE For Limited Time         Attempt Free Tests at GyanPal.com

CS, EC, EE, IN and ME branches

Permutation and Combination Practice Problems :Aptitude Questions

ABOUT THE PAGE


PageInfo


This page is on the topic of "Permutation and Combination" as a subtopic of Aptitude Test Questions and Answers. To do well in aptitude questions one need to be confidant in Permutation and combination. Practicing aptitude problems online on Permutation and Combination topic will prove to be very helpful for those who plan to crack competitive exams like BANK PO, Bank Clerk, various Teaching Exams etc.

Instructions
  • Attempt all question.
  • Click on options to check your answer.
  • Use the timer on the right if want to calculate time taken in solving the questions.


Common information for questions numbered 1 to 4

There are 5 freshmen, 8 sophomores, and 7 juniors in a chess club. A group of 6 students will be chosen to compete in a competition.

1.

Common Data Question 1/4

How many combinations of students are possible if the group is to consist of exactly 3 freshmen?

A.

5000

B.

4550

C.

4000

D.

3550

View Ans

 

Answer – (B)

Solution:

Here we need the number of possible combinations of 3 out of 5 freshmen,${^ 5C_3}$, and the number of possible combinations of 3 out of the 15 sophomores and juniors, ${^ 15C_3}$.

Note that we want 3 freshmen and 3 students from the other classes.

Therefore, we multiply the number of possible groups of 3 of the 5 freshmen times the number of possible groups of 3 of the 15 students from the other classes.

${^ 5C_3} × {^ 15C_3} =$ 4,550

2.

Common Data Question 2/4

How many combinations of students are possible if the group is to consist of exactly 3 freshmen and 3 sophomores?

A.

360

B.

460

C.

560

D.

660

View Ans

 

Answer – (C)

Solution:

This second part of the problem is similar to the first except that the choice of the second group of 3 comes only from the sophomores.

Again we want 3 freshmen and 3 sophomores so we multiply the number of groups of freshmen times the number of groups of sophomores.
${^5C_3} × {^8C_3} =$ 560 

3.

Common Data Question 3/4

How many combinations of students are possible if the group is to consist of an equal number of freshmen, sophomores, and juniors?

A.

5880

B.

6880

C.

7880

D.

8880

View Ans

 

Answer – (A)

Solution:

An equal number of students from each of the three classes mean 2 students from each class, that is, 2 freshmen and 2 sophomores and 2 juniors.
${^5C_2} × {^8C_2} × {^7C_2}=$ 5,880

4.

Common Data Question 4/4

How many combinations of students are possible if the group is to consist of all members of the same class?

A.

20

B.

25

C.

30

D.

35

View Ans

 

Answer – (D)

Solution:

This part of the problem seems similar to the first three parts but it is very different.

Here we want 6 freshmen or 6 sophomores or 6 juniors.
The group cannot be all freshmen since there are only 5 freshmen.
Therefore, the group can be a group of 6 sophomores or 6 juniors.

Number of groups of sophomores $= {^8C_6} = 28$
Number of groups of juniors$ = {^7C_6} = 7$

Since we want the number of groups of 6 sophomores or 6 juniors, we want the sum of each of these possibilities:
$= 28 + 7 =$ 35

5.

How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?

A.

256

B.

24

C.

12

D.

None of these

View Ans

 

Answer – (B)

Solution:

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4.

Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)

Finally B4 will receive the the last remaining prize (so 1 way)
So total ways would be: $4*3*2*1=\bo 24$ Ways

Hence, the 4 prizes can be distributed in $\bo 24$ ways.