I'm learning about the multinomial theorem and working 2 examples in a book. I thought I understood the examples until I did example 5c. I don't understand why these two examples are different. In example 5c it says that order is now irrelevant. What do you mean order is irrelevant? Why was the order relevant in 5b? In both examples we are separating into 2 groups of 5... Thank you in advance!!!

EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?

Solution is 10! / (5! 5! ) = 252

EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?

Solution. Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is

[10! / (5! 5!) ]/2!

  • $\begingroup$ OK I think I understand now. When it's talking about ordering its referring to the whole team and not the actual teammates. So in example 5B, they are distinguishing between Team A: ABCDE and Team B: FGHIJ compared against Team A: FGHIJ and Team B: ABCDE. But in Example 5C it doesn't matter which Team A or Team B is assigned! $\endgroup$ – user1527227 Apr 18 '13 at 17:37
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    $\begingroup$ And the given solution to 5b is... WRONG. It is $\binom{10}{5} = 10! / 5! 5! = 252$. Typo(es)? $\endgroup$ – vonbrand Apr 18 '13 at 19:32
  • $\begingroup$ Yeah sorry it was a typo. thanks for this fix. $\endgroup$ – user1527227 Apr 19 '13 at 18:34

Let us take a much smaller example, $4$ people. We want to divide them into two groups of two each, one group to wear nice blue uniforms, the other to wear brown and yellow stripes. How many ways are there to do the division? We need to choose who will wear the blue uniforms. This can be done in $\binom{4}{2}=6$ ways.

Now consider the ways to divide them into two groups of two each, no uniforms. Call the people $a$, $b$, $c$, and $d$. As soon as we decide who goes with $a$, we will have done the division. So there are $3$ ways to do the job.

Another way of thinking about the second problem is that we first divide the people into two groups-with-uniform. Then we take away the uniforms. The two old divisions $a$ and $c$ wear blue, $b$ and $d$ wear browm/yellow stripes and $a$ and $c$ wear brown/yellow stripes, and $b$ and $d$ wear blue now become a single division into two groups. So to count the number of divisions into uniormless groups, we divide $\binom{4}{2}$ by $2$.

Remark: The idea generalizes. For example, take $20$ people, and divide them into $4$ groups of $5$ people each, one group to wear blue, another white, another red, and another black. First choose who will wear blue. This can be done in $\binom{20}{5}$ ways. For every way of doing this, there are $\binom{15}{5}$ ways to decide who will wear whie, and then $\binom{10}{5}$ ways to decide who will wear white, for a total pf $\binom{20}{5}\binom{15}{5}\binom{10}{5}$.

This can also be written as the multinomial coefficient $\binom{20}{5,5,5,5}$.

Now how many ways are there to divide them into $4$ uniformless groups? When people take off their uniforms, $4!$ uniformed divisions collapse into one, so we need to divide by $4!$.

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    $\begingroup$ You are welcome. In combinatorial problems, it can be useful to look at a small concrete example where one can count explicitly. This can provide a useful check. $\endgroup$ – André Nicolas Apr 18 '13 at 17:52
  • $\begingroup$ This is a great explanation. Wonderfully explained. $\endgroup$ – Sherlock Watson Dec 13 '18 at 14:11

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