Exercise 6. We have 102 distinct gifts. We want to distribute them to 100 winners of a contest, so that each winner receives at least one gift. Let $N$ be the number of ways to do this. Calculate

$$
\frac{N \times 48}{1 \times 2 \times 3 \times \cdots \times 102}
$$
Solution to Exercise 6 First case: one of the winners receives 3 gifts (so we need to choose 3 gifts out of 102, then 1 person out of 100, and then assign 99 gifts to 99 people). This results in

$$
\frac{102 \times 101 \times 100}{6} \times 100 \times 99!=\frac{100 \times 102!}{6}
$$

possibilities.

Second case: two winners each receive 2 gifts. We select 2 people out of 100, then 2 gifts out of 102, then 2 gifts out of the remaining 100, and finally we need to assign 98 gifts to 98 people. This results in

$$
\frac{100 \times 99}{2} \times \frac{102 \times 101}{2} \times \frac{100 \times 99}{2} \times 98!=102!\times \frac{100 \times 99}{8}
$$

In total, this gives

$$
102!\times\left(\frac{100 \times 99}{8}+\frac{100}{6}\right)=102!\times 100 \times \frac{99 \times 6+8}{48}
$$

so \( N=100 \times(99 \times 6+8)=100 \times(100 \times 6+2)=60200 \).