100 people in 20 rooms

Question

I have to put 100 people in 20 different rooms so that no room is empty and the president is alone in a room. Can I first from 100 people choose 1 to be the president and then from the remaining 99 people to choose 19 because they can't be in the room where the president is?

Answer 1

After assigning the president to his room ($20$ choices), we have to assign the remaining $99$ people to the remaining $19$ rooms so that no room is empty.

This is a PIE problem as Mike Earnest has pointed out.

Let $A_i$ be the set of arrangements in which the $i$th room is empty and $|A_i|$ be its size.

Then $$|A_1 \cap A_2 \cap \cdots \cap A_k|=(19-k)^{99}$$

By PIE, the number of arrangements in which at least one room is empty is given by

$$\sum_{k=1}^{19}(-1)^{k-1}\binom{19}{k}(19-k)^{99}$$

Thus the number of arrangements in which NO room is empty is:

$$19^{99}-\left [\sum_{k=1}^{19}(-1)^{k-1}\binom{19}{k}(19-k)^{99} \right]$$

$$=\sum_{k=0}^{19}(-1)^k\binom{19}{k}(19-k)^{99}$$

Thus the final answer is

$$20 \times \sum_{k=0}^{19}(-1)^k\binom{19}{k}(19-k)^{99}$$