Understanding counting using multinomial coefficients

Question

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:

1. 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 $$\frac{10!}{5!2!3!}$$

2. 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 $$\frac{10!}{5!5!}$$

3. 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.

Answer 1

Maybe it's easier to see with a simpler example:

2 kids, Amanda and Ling, are to be divided into two teams A and B of size 1 ...

This is counting these combinations separately:

• Team A: {Amanda}; Team B: {Ling}.
• Team A: {Ling}; Team B: {Amanda}.

Thus we get $$\frac{2!}{1!\ 1!} = 2$$ possibilities.

2 kids, Amanda and Ling, divide themselves up into two teams of 1 to play basketball at the playground

This is counting both possibilities above as the same thing. Or in other words, it only counts this possibility:

• Team: {Amanda}. Team: {Ling}.

Thus we get $$\frac{2!}{1!\ 1!} / 2! = 1$$ possibility.

Now just replace 2 with 10, and 1 with 5.

Answer 2

The sets patrol, station, reserve in example 1 are clearly distinct. Thus the ordering among these sets matter.

The teams A and B in example 2 are clearly distinct, they are given names. It makes a difference whether someone is in Team A or Team B.

In example 3, it doesn't matter what you call the teams, so the order doesn't matter.

Answer 3

The distinction is best seen as one between colouring and partitioning a (finite) set (possibly subject to some restrictions). A colouring is simply the association of a colour (presumably from a given palette) to each element. On the other hand a partitioning just defines a notion of "togetherness" or "being on the same team" holding within certain subsets (called classes of the partitioning); technically this is an equivalence relation on the set. There is nothing external to distinguish the classes; there are just its members that can be used to identify a class. So every colouring defines a partitioning (into classes consisting of elements having a common colour), but a partitioning does not determine a single colouring, since going from a partitioning to a colouring we are free to choose a colour to identify each class with (as long as distinct classes get distinct colours).

In problems 1 and 2 there are the palettes { patrol, station, reserve } respectively $\{A,B\}$ that can be considered to define the colours to be used, but in 3. there is no such thing: each team is just defined by its members. It we had say red and blue shirts to identify the teams, then we would have been talking about a colouring, and the answer would be twice as large, but the actual question is about a partitioning (into $2$ classes of size$~5$).