Question

Combinatorics:
In how many variations can you sit 10 people on two benches if one has 5 seats and the other has 7 seats?

Answer 1

I'm not sure this is correct, but here's my intuition. No matter where the 10 people sit, there will always be 2 empty seats. Use 10 choose 2 to identify all the possible combinations of empty seats. Then for the remaining 10 seats, there are 10! ways to seat 10 people in 10 seats. Thus,

\[\left(\begin{matrix} 10 \\ 2\end{matrix}\right) 10!\]

Answer 2

The textbook answer is 239500800, what you proposed seems plausible but does not match the answer

Answer 3

I've tried all kinds of combination functions but just can't seem to nail it.

Answer 4

Finding all possible instances and then removing the two empty seats could be it, but I'm not exactly sure how to put it into numbers

Answer 5

Got it. It's 66 * 10!

Answer 6

\[\frac{ 12! }{ 2! }\] seemed to give the answer, now the question is if it's coincidental or actually the way.

Answer 7

Dividing by 2! reminds me the formula of "getting rid of unwanted combinations" in identical objects problems.

Answer 8

Instead of 10 choose 2 for the empty seats, you have to take into account that you have 2 different benches. So the better way to choose the empty seats is by
\[\left(\begin{matrix}7 \\ 1\end{matrix}\right) \left(\begin{matrix}5 \\ 1\end{matrix}\right)+\left(\begin{matrix}7 \\ 2\end{matrix}\right) + \left(\begin{matrix}5 \\ 2\end{matrix}\right)\]
Where the first term is if you have an empty seat in each of the benches, then 2 empty seats in the 7 person bench, then 2 empty seats in the 5

Answer 9

\[\frac{ n! }{ n_{1}!\times n _{2}!\times n _{3}!... }\]

Answer 10

I think your way also makes sense? It's a little less intuitive to me though.

Answer 11

The multinomial coefficient doesn't seem necessary... at least I can't figure out any initial set up that would reduce to \(\dfrac{12!}{2!}\).

\(\dfrac{12!}{2!}\) only seems to count the number of ways 10 people (plus two empty slots) can be arranged, regardless of how they are partitioned among the benches, and the empty slots are repeats hence we divide by 2!.

Does this seem reasonable?