A basic but a bit annoying combinatorics problem

ernie

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How many mixed hockey teams can be formed from six married couples, three spinsters and one bachelor if no wife will play without her husband? (A mixed hockey team = either five men and six women or six men and five women)
 

Dr.Peterson

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Since the word "annoying" implies that you have made some attempts, how about showing us some of what you have tried, so we can know where you need help? That's what we ask in our submission guidelines.

It looks to me like a problem that has to be broken into cases. I'd probably start with one of the two main cases, say 5 men and 6 women. Within that case, you might have to separately count the cases with 1 wife, 2 wives, 3 wives, 4 wives, 5 wives, and 6 wives. Yes, that sounds like a lot of cases; but maybe as you proceed, you'll find a shortcut. (I think, for example, that some of my list of cases will evaporate entirely.) That's another reason to want to see what you've tried.
 

pka

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How many mixed hockey teams can be formed from six married couples, three spinsters and one bachelor if no wife will play without her husband? (A mixed hockey team = either five men and six women or six men and five women)
This is indeed annoying, but also fun. Note there are only three unmarried women.
That means we must use a minimum of two couples and a maximum of five couples( you may want to think about that last point).
I would consider four cases: \(\displaystyle \#C=2,~3,~4, \text{ or }5\) couples used.

Let type I=five men and six women and type II=six men and five women; selecting couples, spinsters, & other men.
Here is an example: \(\displaystyle \#C=3~\&~\text{type }I.\) \(\displaystyle \binom{6}{3}\binom{3}{3}\binom{4}{2}\)
while \(\displaystyle \#C=3~\&~\text{type }II.\) \(\displaystyle \binom{6}{3}\binom{3}{2}\binom{4}{3}\)
 

ernie

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Thanks! I had thought it exactly as pka suggests: number of teams with 2 of the married women, with 3, with 4 with 5; then add all this up. With 2 married women you can only form 6C2 x 3C3 x 5C4 teams. With 3, just as pka suggested. With 4, it's 6C4 x (3C2 x 3C1 + 3C1 x 3C2). With 5: 6C5 x (3C1 x 5C0 + 1C1). All this should add up to 876, but I do not get THAT number!
 

pka

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Thanks! I had thought it exactly as pka suggests: number of teams with 2 of the married women, with 3, with 4 with 5; then add all this up. With 2 married women you can only form 6C2 x 3C3 x 5C4 teams. With 3, just as pka suggested. With 4, it's 6C4 x (3C2 x 3C1 + 3C1 x 3C2). With 5: 6C5 x (3C1 x 5C0 + 1C1). All this should add up to 876, but I do not get THAT number!
May I ask where you got the number 876?
As I said above there are two types: \(\displaystyle I, \text{ 6 women & 5 men }~\&~II\text{ 6 men & 5 women }\)
\(\displaystyle \text{ type } I~:\) \(\displaystyle \sum\limits_{k = 3}^5 \binom{6}{k}\binom{3}{6-k}\binom{7-k}{5-k}=273\) SEE HERE

\(\displaystyle \text{ type } II~:\) \(\displaystyle \sum\limits_{k = 2}^5 \binom{6}{k}\binom{3}{6-k}\binom{7-k}{6-k}=462\) SEE HERE

That is \(\displaystyle 273+462=735\) I do not see where my under-count is mistaken.
We cannot use all six couples due to the six/five makeup of the teams.
 

ernie

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Neither do I see any undercount. 876 though is the answer provided by Backhouse & Houldsworth's Pure Mathematics, where this problem is from.
 
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