In how many ways can 3 prizes be distributed among 5 students when

Question

In how many ways can 3 different prizes be distributed among 5 students when a student may get any number of prizes?

(A) 60
(B) 100
(C) 120
(D) 124
(E) 125

Bunuel · Answer

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Bunuel · Answer

__________________________
Anyone want to try this one?

ramlala · Answer

Prices: 5 x 5 x 5 = 125

IMO E

Mugdho · Answer

KarishmaB  help me plz

Why can't we think this way....Any student can choose any number of prizes... so one can choose 0-3 prizes that is one has 4 options to choose. first student has 4 ways to choose.. second student has also 4 ways to choose and so on..so ways are: 4*4*4*4*4....why this thinking is wrong?

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KarishmaB · Answer

The available choices for the second student are dependent on what first student chose. The available choices for third student depends on what first two students chose etc. So it is not 4*4*4*4*4.

If first student chooses no prize, second student has 4 options (0/1/2/3). 
If first student chooses 1 prize, second students has 3 options (0/1/2). 
etc. 
Also, in this case, it is possible that no student may claim any prize which is not acceptable.

All 3 prizes need to be distributed. For each prize, we must choose 1 of 5 students.

Catman · Answer

Prizes will goes to student, so every prize can go to 5 different student.
If prizes be P1,P2,P3
than for students S1,S2,S3,S4,S5.
Prize P1 can be given to any of the 5 students.
Prize P2 can be given to any of the 5 students.
Similarly Prize P3 can be given to any of the 5 students.
Hence total choices will be 5*5*5= 125.

Posted from my mobile device

rito54 · Answer

Hello! I don't get the solution. Can one student get 2 prizes or am I reading badly the statement?

Bunuel · Answer

Check the highlighted part:

In how many ways can 3 different prizes be distributed among 5 students when  a student may get any number of prizes ?

Soumya_Pinjala · Answer

So I tried to solve it as
3 prizes go to diff people = 5*4*3
2 prizes go to same person and 1 to another = 5*1*4
3 prizes go to same person = 5*1*1

Adding up gives 85. But this is not the answer. Am I missing cases or is this approach wrong. I would like to understand to avoid mistake in the future

Bunuel · Answer

You undercounted the middle case.

If 2 prizes go to one student and 1 to another, you must also choose which 2 of the 3 different prizes go together. That gives an extra factor of 3 (3C2 = 3). So:
 
5 * 4 * 3 + 5 * 3 * 4 + 5 = 125

egmat · Answer

Y ou're undercounting in Case  2 .

Case 1:  All  3  prizes to different people =  5  ×  4  ×  3  =  60 . This is correct.

Case 2:2  prizes to one person,  1  to another. Here's where the mistake is. You wrote  5  ×  1  ×  4  =  20 , but you forgot that the prizes are DIFFERENT. You need to first choose WHICH  2  of the  3  prizes go to the same person.

For example, giving prizes {A, B} to Student 1 and prize {C} to Student 2 is a DIFFERENT outcome than giving prizes {A, C} to Student 1 and prize {B} to Student 2.

So the  correct count  is:
- Choose which  2  prizes go together:  C(3,2)  =  3  ways
- Choose who gets the pair:  5  ways
- Choose who gets the remaining prize:  4  ways
- Total =  3  ×  5  ×  4  =  60

Case 3:  All  3  prizes to same person =  5 . This is correct.

Adding up:  60  +  60  +  5  =  125 . That matches answer  E !

Your missing piece was the  3  ways to select which pair of prizes goes together in Case  2 . That accounts for exactly the  40  you were short ( 60  -  20  =  40 ).

Answer: E

In how many ways can 3 prizes be distributed among 5 students when

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In how many ways can 3 prizes be distributed among 5 students when

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