Question

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

How many ways have at least two students getting nothing?

Answer 1

There are $\binom{4}{2}$ ways to choose 2 student that get nothing. Each prize can be given to one of 2 remaining students. So total number of variants equals to $2^{8} - 2$ (2 variants when one of the students gets all prizes. So answer for the first question is

$$\binom{4}{2} (2^{8} - 2) = 6 \cdot 254 = 1524$$

To find the answer for the second question we need to add number of ways where 3 students get nothing that equals to 4 (when one of the students gets all prizes). So the answer is

$$1524 + 4 = 1528$$