# Permutations, Combinations, and Counting

A group of 63 people are camping together. They have two 6-person tents, three 4-person tents, five 3-person tents, and three 2 person tents. 18 people will sleep outside of the tents under a tarp.

Determine the how many ways the people can choose the tents to sleep. (All the tents are different)

Suppose exactly 4 people snore. Count the number of ways of assigning the people to the tents so that all the snorers share their tents only with other snorers

• What are your thoughts on the problems? Jul 7, 2014 at 19:28
• First thing I did was 63C18 for the different possible ways to sleep outside. Now remains 45 people. I'm trying to think of this problem as 45 balls, with different bins, where each bin has a size restriction. I don't know how to translate how to choose the bins so they add up to at least 45. Jul 7, 2014 at 19:34

The calculation mentioned in the comment began well. Note that the numbers mentioned for the various tents and the tarp add up to $63$, so the tents will be all full.

The people for the tarp can be chosen in $\binom{63}{18}$ ways.

For each such way, the $6$ people in big tent Number 1 can be chosen in $\binom{45}{6}$ ways.

For each such way, the $6$ people in big tent Number 2 can be chosen in $\binom{39}{6}$ ways.

Continue.

When we multiply, and simplify the binomial coefficients, we get something that begins like this: $$\frac{63!}{18!45!}\frac{45!}{6!39!}\frac{39!}{6!33!}\frac{33!}{4!29!}\cdots.$$ There is massive cancellation, and we end up with $$\frac{63!}{18!6!6!4!4!4!3!3!3!3!3!2!2!2!}.$$ You may recognize this as a multinomial coefficient.

Another way of thinking about the problem is that we will line up the people in order of Student Number, or height. Then the assignment to tarp/tents will be made by writing down a $63$-letter word, using $18$ copies of the letter $T$ (tarp), $6$ copies of the letter $A_1$ (big tent Number 1), $6$ copies of the letter $A_2$ (big tent Number 2), and so on.

For the snoring restriction, the snorers will be (i) in one of the $4$-person tents ($3$ choices) or (ii) in two of the $2$-person tents. For Case (ii), the tents can be chosen in $\binom{3}{2}$ ways, and once the tents have been chosen, the people can be assigned to the two tents in $\binom{4}{2}$ ways.

For each of Cases (i) and (ii), we can calculate the number of ways the rest of the people can be assigned by using the method of the first question. Now put the pieces together.

Remark: The problem becomes more complicated if we assume that tents of the same size are indistinguishable.

• Ah thank you, I understand now. I didn't realize that the occupancy of the tents in total was 45. Say for example, the problem changed where the amount of tents doubled (so occupancy changed from 45 to 90) and all the tents were still different. Would I go about solving the problem as you described, or would it become a whole different problem? Jul 7, 2014 at 20:09
• If available tent space goes to $90$, say by doubling the number of tents of each type, the problem becomes vastly more complicated. Jul 7, 2014 at 20:12