I'm studying Chapter 1 of Ross A First Course in Probability Theory (8th Edition) and I'm grappling with multinomial coefficients. All given examples come from this chapter. Specifically $${n \choose n_1 … n_r}=\frac{n!}{n_1! … n_r!}$$ gives the number of ways to choose groups of objects of sizes $n_1, …, n_r$ where $\sum_{i=1}^{i=r} n_i = n$. Then we have the following 3 examples:
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10 officers are to be divided as follows. 5 on patrol, 2 at the station and 3 in reserve. In how many ways can this be done? The answer is of course $$\frac{10!}{5!2!3!}$$
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10 kids are to be divided into two teams A and B of size 5 each where each team will play in a separate division. In how many ways can this be done? Again, the answer is similar to what we'd expect $$\frac{10!}{5!5!}$$
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10 kids divide themselves up into two teams of 5 to play basketball at the playground. In how many ways can this be done. This is where my confusion begins.
The example says the answer is $$(\frac{10!}{5!5!})/2!$$ because even though this looks like the previous problem, it is different since the order doesn't matter here.
Firstly, it looks exactly the same. I do not see why order matters in EITHER of examples #$2$ and #$3$. Secondly, example #$1$ looks exactly like the situation of example #$2$ except with $3$ groups instead of $2$ so, if order mattered in example #$2$, then it should have mattered in example #$1$, no?.
So, my question: What am I missing here? Any feedback is much appreciated.
Best Answer
Maybe it's easier to see with a simpler example:
This is counting these combinations separately:
Thus we get $$\frac{2!}{1!\ 1!} = 2$$ possibilities.
This is counting both possibilities above as the same thing. Or in other words, it only counts this possibility:
Thus we get $$\frac{2!}{1!\ 1!} / 2! = 1$$ possibility.
Now just replace 2 with 10, and 1 with 5.