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I ran into a problem in a textbook and though the solution in the book makes sense, I don’t know why my solution of it wouldn’t work.

So the question is about how there are 20 people, which are 10 groups of married people. The question asks how do we determine the number of ways that we can select 5 people out of the twenty, making sure no 2 people in that group of 5 are married.

The answer was to first select five groups of the ten married groups, then pick 1 person of the five married groups to achieve the result: $$\binom{10}{5}\binom{2}{1}^5$$

I just don’t understand how they can use the binomial coefficient $\binom{10}{5}$ to pick the group of married couples.

How come we can’t split the $20$ people into two groups: the husbands and the wives? Then we choose one group of those two, then choose five of the ten people left. That would result to: $$\binom{2}{1}\binom{10}{5}$$ which is incorrect. What am I failing to understand with my answer?

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 26 '17 at 9:48
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By choosing five of the ten married couples, we ensure that no two people we select are from the same couple. For each such selected couple, we have the option of selecting either the husband or the wife, which is why the number of permissible selections is $$\binom{10}{5}\binom{2}{1}^5$$

Where did you go wrong in your attempt?

For simplicity, let's assume that there are $10$ opposite sex couples. By selecting the husbands or the wives, then selecting five people from the selected group, you only permit selections of five people of the same sex. However, the way the problem is stated we may select both men and women provided that our selection does not include both members of a married couple.

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    $\begingroup$ Oh I understand, so because I began with $\binom{2}{1}$ which selected one group of the two sexes, this limits the range of groups I can select in total because it will only ever be a group of the same sex, and the problem never asked for such a restriction. Is that right? $\endgroup$ – R. Carton Oct 26 '17 at 21:32
  • $\begingroup$ Yes, that is what I meant. $\endgroup$ – N. F. Taussig Oct 26 '17 at 22:05
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You assumed that each person has a tag, "husband" or "wife", and you also know which tag each one is attached to.

But these are not given in the problem.

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aliter

  1. Choosing 1 from 20
  2. Choosing 1 from 18 since (1 is chosen already and partner is avoided)

So on$\ldots$$$\frac{20*18*16*14*12}{5!}$$

  1. Since order doesn't matter it's divided by $5!$
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