The question states that:
9 players are playing a game and they have to gather at 3 identical booths. How many
ways can this be done if each booth must have at least 2 players if the arrangement of the
players does not matter?
Personally, I have tried finding any general formula. I found Stirling number of the second kind but it can only be applied to non-empty groups. In this problem that I am facing, it clearly states that it requires at least 2 players in each booth and hence I am at a loss of how to solve this case, and if possible for my own learning, find a generalised formula.
My solution is that
There can be $3$ ways to arrange $9$ players into $3$ groups with each group having more than $2$ members.
$$\text{Case 1: AA BB CCCCC (2, 2, 5 combination)}$$
$$\text{Case 2: AA BBB CCCC (2, 3, 4 combination)}$$
$$\text{Case 3: AAA BBB CCC (3, 3, 3 combination)}$$
$$\text{Case 1: } \frac {{9 \choose 5} \times {4 \choose 2} \times {2 \choose 2}} {2!} = 378$$
$$\text{Case 2: } {{9 \choose 4} \times {5 \choose 3} \times {2 \choose 2}} = 1260$$
$$\text{Case 3: } \frac { {9 \choose 3} \times {6 \choose 3} \times {3 \times 3}} {3!} = 280$$
Total number of possible ways: $378 + 1260 + 280 = 1918$
I was wondering if this solution is correct. If this solution is correct, is there any generalised formula I could use?
Best Answer
The answer seems to be $1918$, as you found. Here is another way to find it, though there may be a simpler way. Initially order the booths $1$, $2$, and $3$. Then condition on how many people are in booths $1$ and $2$, choose the people going in booths 1,2, and 3, and finally divide by $3!$ since order of booths doesn't matter. This way of thinking leads to the solution
$$\frac{1}{6}\sum_{x=2}^5\sum_{y=2}^{7-x}\frac{9!}{x!y!(9-x-y)!} = 1918.$$