combination gre problem # 24

Question

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

Can anyone explain me why this answer is wrong: 4*4*4 ?

Answer 1

Let 4 prizes, $p_1, p_2, p_3, p_4$ and 3 boys, $b_1, b_2, b_3$

Each price $p_i$ has a choose $c_i$ of a boy. So

$c_1 \in \{b_1, b_2, b_3\}$
$c_2 \in \{b_1, b_2, b_3\}$
$c_3 \in \{b_1, b_2, b_3\}$
$c_4 \in \{b_1, b_2, b_3\}$

And set of the options is $\{b_1, b_2, b_3\} \times \{b_1, b_2, b_3\} \times \{b_1, b_2, b_3\} \times \{b_1, b_2, b_3\}$,
and the size of the set is $|\{b_1, b_2, b_3\}|^4 = 3^4$.

Answer 2

You only have one of each prize, so it is true that you have $4$ possibilities for the first boy. For the second boy, you only have $3$ prizes left, since you have already given $1$ away. Likewise, there are $2$ possibilities for the third boy, giving you a total of $4\cdot 3\cdot 2=24$.

EDIT: I assume you have to give each boy a prize. If each of the $4$ prizes are given randomly (such that a boy might win more than one), you would have $3^4$ different ways.

Answer 3

Since each boy is eligible for all the prizes;there are 3 possibilities for each prize, giving us 3^4=81 possibilities.