Workbook question:
Professor is giving an exam. (A) How many ways are there to assign five proctors to the three exam rooms, so that each exam is proctored by at least one person?
Answer:
$n = 5, a = 3$
$3^5 - \binom{3}{1}(3-1)^5 + \binom{3}{2}(3-2)^5 = 150$
I understand the answer in terms of the formula, but I don't understand the answer in terms of the principle, or in any "common sense" way. I can't visualize how the Inclusion-Exclusion principle applies to this sort of situation.
There are $3^5$ ways to distribute the five proctors to the three rooms. But some of the distributions leave a room unsupervised: for each of the $\binom{3}{1}$ rooms, there are $(3-1)^5$ ways to distribute the five proctors to the other two rooms, we're at $3^5 - \binom{3}{1}(3-1)^5$ now. But the distributions where two rooms were left unsupervised are subtracted for each of the unsupervised rooms, so twice. Thus we must correct and add the number of distributions that leave two rooms unsupervised. There are $\binom{3}{2}$ combinations of two rooms, and for each of these, there are $(3-2)^5$ ways to distribute the five proctors to the remaining rooms. Since with three rooms and more than $0$ proctors, there is no way to leave three rooms unsupervised, so we can stop now, at $3^5 - \binom{3}{1}(3-1)^5 + \binom{3}{2}(3-2)^5$.