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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.

RobPratt
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Salazar_3854708
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3 Answers3

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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.

Rebecca J. Stones
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  • Hi Rebecca, thank you very much for your response. While it is quite clear, I've seen another question where "12 people are to be divided into 3 groups of sizes 3, 4 and 5" and the answer given is the one that DOES take order into account. But to me, the language doesn't make a distinction between the groups so I think the order shouldn't matter here. Do you agree with my assessment of this? – Salazar_3854708 Jun 18 '22 at 12:20
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    I agree, the phrasing isn't fully unambiguous. In practice, you often have to figure out the most likely interpretation of an author's wording (labelled vs. unlabelled, distinguishable vs. indistinguishable). For distinct group sizes (e.g., 3, 4, and 5), it can only mean "distinguishable" since we can distinguish the groups by their size (e.g., we can't swap a team of size 3 with a team of size 4 and consider it the same thing). But for non-distinct group sizes (5 and 5) it could be either depending on the wording of the problem. – Rebecca J. Stones Jun 18 '22 at 12:27
  • This is immensely helpful Rebecca. I appreciate the clarification. – Salazar_3854708 Jun 18 '22 at 13:04
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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.

kodlu
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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$~5$).

Marc van Leeuwen
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