Aptitude 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.**

**Vishal**

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**Piyush Tripathi**

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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**

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

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

**Sanchit**

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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

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

**Satish Kumar**

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yeah the solution provided is in contradiction to the question asked.

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**

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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**

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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 $3\times 3\times 3\times 3=81.$

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.

Had 1 person is allowed to receive multiple prizes, your solution could have been right.

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**

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

**Jessie**

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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**

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

**Ankt**

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

**Antonio Banderras**

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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.

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:

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

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