# Easy Permutation-Combination Solved QuestionAptitude Discussion

 Q. How many ways can 4 prizes be given away to 3 boys (one boy can receive 1 prize), if each boy is eligible for all the prizes?
 ✖ A. 256 ✔ B. 24 ✖ C. 12 ✖ D. None of these

Solution:
Option(B) is correct

Let the 3 boys be $B_1, B_2, B_3$ and 4 prizes be $P_1, P_2, P_3$ and $P_4$.

Now $B_1$ is eligible to receive any of the 4 available prizes (so 4 ways)

$B_2$ will receive prize from rest 3 available prizes(so 3 ways)

Finally, $B_3$ will receive his prize from the rest 2 prizes available(so 2 ways)

So total ways would be: $4 \times 3 \times 2 =\textbf{24}$ Ways

Hence, the 4 prizes can be distributed in $\textbf{24}$ ways.

## (17) Comment(s)

Vishal
()

why cannot we take 4C3 here? how it is not a permutation question? plzz anyone tell

Piyush Tripathi
()

Guys, when an award is given to any of the students then next student has only three choices.

So, the answer is 24 only.

Abhinav Tyagi
()

there are 3 boys in question. is there a mistake in solution?

Matt
()

I guess they have made the correction, it's not there now.

Sanchit
()

answer is $3^4 = 81$ as every award has 3 choices of student and all of them are eligible. (also 1 boy is more in solution) the answer they posted is when 4 students receives 4 awards one each

Ashutosh
()

This is incorrect method as 1 boy should not get more than 1 prize. Correct answer should be 24 (4x3x2).

Satish Kumar
()

the question tell that 3 boys are eligible for all 4 prizes (practically illogical)

so we have 3 boys A B C and 4 prizes 1 2 3 4

So, A is elligible for all 1 2 3 4

B is again eligible for all 1 2 3 4

C is again eligible for all 1 2 3 4

so we have

4 * 4 * 4

A B C

$4^3 = 64$ ways

Vikas
()

three student can get by 3! ways=6

and there can be another 4 ways as 4 prizes avail

So, total $3!*4=24$

Sudarsan K
()

No way!

If B1, B2 & B3 are boys and P1, P2, P3 & P4 are the prizes, then P1 can be given not only to B1, but can also be given to B2 or B3.

From the prize perspective;

P1 can be distributed to any 1 of the 3 boys, so 3 possibilities.

Similarily the P2, P3 & P4 also have 3 possibilities each.

So the answer should be $3\times 3\times 3\times 3=81.$

David K
()

I have to agree with Sudarsan.

I think that if there are 3 boys and 4 prizes then it is simply 3 raised to the power 4 which is the number of prizes available.

Geetika
()

For example, you are saying that boy $B_1$ is getting all the prizes i.e. $P_1, P_2$ and $P_3$, which I don't think is the desired possibility.

Mini
()

Hey I think ans must be $3^4=81$

Jessie
()

I think the solution is probably wrong.

Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize.

Hence, the 4 prizes can be distributed in $3^4= 81$ ways.

James
()

Hey Relk, how is there a B4 when there are only three boys...

Ankt
()

i thin that all 3 boys can get 4 awards so $4*4*4=64$ could be answer.

Antonio Banderras
()

I don't like this question, mainly because once you give a prize to a boy, that boy should stick with the prize, and the second boy would only have 3 prizes to choose from left.

I believe it should be a combination without repetition, hence my answer is 12.

Relk
()

Hey Antonio, I also feel the existing solution is incorrect. Let me explain with the help of an example:

Let's say boys are B1, B2, and B3 whereas prizes are P1, P2, P3, and P4.

Now B1 is eligible to get any of the 4 available prizes (i.e. 4 choices)

B2 will get from the rest 3 available prizes(i.e. 3 choices)

B3 will get prizes from the rest available 2 prizes (i.e. 2 choices)

So total ways would be:

$4*3*2= \textbf{24 Ways}$