In how many ways can one distribute ten distinct pizzas among four students with exactly two students getting nothing?

Question

In how many ways can one distribute ten distinct pizzas among four students with exactly two students getting nothing? How many ways have at least two students getting nothing?

For the first part I tried to solve it as

$$x1+x2 = 10 \text{, } xi \ge 1$$ Which would give $$ x1+x2 = 8, \text{ } xi \ge0$$

Then I computed $$9\choose 8$$ multiplied by 6.

Why is this wrong? I seen the answer and they are doing something alone the lines of $$ (2^{10} - 2) * 6$$

This relationship is not clear to me, can anyone explain?

Answer 1

Here is one way to obtain this count:

First, choose which students will receive nothing. This can be done in $\binom{4}{2}$ ways.

Second, choose how to distribute the 10 distinct pizzas to the remaining 2 distinct students. For each pizza, you have a 2 choices of which student will receive it. In total, the distribution can take place in $2^{10}$ ways. This is almost correct, but two of these distributions are illegal, namely the one in which student A gets all the pizzas or the one in which student B gets all the pizzas. So there are actually $2^{10} - 2$ legal distributions.

Taken together, we get $\binom{4}{2} \cdot (2^{10} - 2) = 6(2^{10} - 2)$ legal ways to distribute the pizzas under your constraints.

Answer 2

There are 6 ways to pick the students that don't eat. Once we pick them

How many ways are there to give out $10$ pizzas to two students with both getting something?

at first glance $2^{10}=1024$ but we substract the $2$ cases where a student gets $0$,

So the answer is $1022\cdot6=6132$

Answer 3

For the second part, the number of ways for at least two students to get nothing, take the answer from the previous section, and add the number of ways for three students to get nothing, which is $4$ (three students getting nothing means one student getting everything and there are four different students who could get fat everything).

So taking the answer from the second part as given by other users, we get ${{4}\choose {2}} \cdot (2^{10} - 2) + 4 = 6136$