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

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Anyone want to try this one?

IMO Answer = 120

Prices: 5 x 5 x 4 = 100

IMO B

I think the answer should be 120. (5p3 + 3c2*5p2)

5*5*4 cannot be the answer as I believe this situation is not the only situation (for eg. for the first prize lets assume out 5 students, student 1 got it. The second price now is also open to 5 students, lets say anyone apart from student 1 got it. Now the last prize is also open again to 5 students and not 4).

Hope anyone would like to add my explanation.

Posted from my mobile device

Yes.
Understood my mistake.
Your answer makes sense.
It can also be solved rather quickly by subtracting the number of cases when all gifts go to 1 student from the total number of cases.
= (5*5*5 - 5) = 120

All possible distributions: 5x5x5 = 125 (Including students get all three prices)
All distributions where students get all prizes: 5 (Since there are only 3 prizes, only one student can get all three at the same time, since we have 5 students it's 5.)

125-5 = 120

IMO C

Since all 3 prizes (different) cannot be given to one student the distribution will be in the following ways:

2-1-0-0-0
First we will select one student who gets 2 prizes = 5c1 ways, then we will select which 2 prizes he gets = 3c2 ways. And then we will be left with 4 students and one prize, which will be distributed in 4c1 ways.
Total = (5c1*3c2)*(4c1) = 60 ways

1-1-1-0-0
First we will select three students who gets 1 prize each = 5c3 ways, then we will select which prize each student gets = 3c1*2c1*1c1 ways.
Total = 5c3*3c1*2c1*1c1 = 60 ways

We can also select three students first and then permute 3 prizes among them the answer will be same = 5c3*3p3

Total ways = 60+60= 120

Option A

Let the 3 distinct prizes be: X —-Y—-Z

Let the 5 people be: A, B, C, D, and E


Since no one person can receive all 3 prizes, there are 2 ways we can distribute the prizes:

[1–1–1] or [2–1]

(Scenario 1)
Distribute the prizes so that 3 different people each get 1 prize

(1st)out of the 5 people, how many ways are there to choose the 3 people who will get the prize:

“5 choose 3” = 5! / 2! 3! = 10

And

(2nd)for each of the 10 possible groups, we can divide the 3 distinct prizes among 3 different people in:

3! = 6 ways

Scenario 1:

(10) * (6) = 60 ways


Scenario 2:
Distribute the prizes so that 1 person gets 2 of the distinct prizes and another 2nd person gets 1 of the prizes [2–1]

(1st) how many ways are there to choose the 2 people out of the 5 total who will receive the prizes?

“5 choose 2” = 5! / 2! 3! = 10 ways

And

(2nd) how many ways are there to divide the 3 prizes up such that there is one stack of 2 prizes and another stack of 1 prize

“3 choose 2” * “1 choose 1” = 3 ways

And

(3rd)how many ways are there to distribute the two distinct “stacks” of prizes to the 2 people chosen?

2! = 2 ways

Scenario 2:

(10) * (3) * (2) = 60 ways

Answer

60 + 60 =

120 ways

Posted from my mobile device

Total no of options : - 5 x 5 x 5 = 125
Less ways in which all prices receiving by single student: 5

Answer is 125 - 5 = 120

In case you were not quick to pick the "Total Possibilities-Possibilities of one having all 3 trophies" method
You could use the permutation method I used:
Poss. of 3 out of 5 having a trophy each: 5*4*3=60
Poss. of 2 out of 5 having the trophies: 5*4=20, Since there are 3 different trophies they can be distributed in 3 ways (3C2)
So 20*3=60
Add both scenarios= 60+60=120 (C)

By far the best method is 125-5=120 but it didn't click to me as I started working on this. Dammit

Hey Bunuel, can you please guide me where I am going wrong?

The way I approached this was that

1st Prize: 5 ways
2nd Prize: 5 ways
3rd Prize: 4 ways (Since the same person cannot get all prizes)

I understand your approach 125-5 = 120 but what is the incorrect part in mine? That its giving me the incorrect answer?

Thanks

Say the student are A, B, C, D and E, and prizes are x, y, and z. Now, if x and y prizes go to different students, then z will have 5 options not 4.

A -- B -- C -- D -- E
x -- y

z can go to anyone from the 5.

So, your method will give smaller number of combinations than there actually is.