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

A. 81
B. 64
C. 48
D. 27
E. 12


Since each prize can have 3 choices (a boy being a choice), the 4 prizes can be given to 3 boys in
3 x 3 x 3 x 3 = 3^4 = 81 ways.

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

A. 81
B. 64
C. 48
D. 27
E. 12


Since each prize can have 3 choices (a boy being a choice), the 4 prizes can be given to 3 boys in
3 x 3 x 3 x 3 = 3^4 = 81 ways.

Answer: A

Why can't we say each boy can have 4 choices and then do 4x4x4
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Bunuel
How many ways can 4 different prizes be given away to 3 boys, if each boy is eligible for all the prizes?

A. 81
B. 64
C. 48
D. 27
E. 12


Since each prize can have 3 choices (a boy being a choice), the 4 prizes can be given to 3 boys in
3 x 3 x 3 x 3 = 3^4 = 81 ways.

Answer: A

Why can't we say each boy can have 4 choices and then do 4x4x4

Because the question is about assigning 4 prizes to 3 boys, each prize must go to one boy. So each prize has 3 choices, giving 3^4 = 81 ways.

If you flipped it and said each boy has 4 choices, you'd end up assigning multiple boys to the same prize, which makes no sense, since each prize can only go to one person.
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Bunuel So one boy can have all the prizes while others have none?
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Bunuel So one boy can have all the prizes while others have none?
Yes, in the original question, 3^4 = 81 represents all possible ways to assign each of the 4 prizes to any of the 3 boys. That includes cases where one boy gets all 4 prizes and the others get none.
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