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Problem: Find the number of ways which we can divide ten people into five groups, each containing two person.

I said we set the first group ($10\times 9$), the second group ($8\times 7$), ... and the whole work is done in $10!$ ways. But the book's solution says we randomly choose a person and he can choose another for his team in 9 ways and so on, and finally the whole work is done in $9\times7\times5\times3$ ways. What is wrong with my solution?

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3 Answers 3

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Your answer $10!$ counts the number of ordered groups of ordered pairs, but the question asks for unordered groups of unordered pairs. This results in two problems:

  • These groups are counted separately: $$\big\{ \{\color{blue}1,\color{blue}2\}, \{3,4\}, \{5,6\}, \{7,8\}, \{9,10\}\big\}$$ $$\big\{ \{\color{blue}2,\color{blue}1\}, \{3,4\}, \{5,6\}, \{7,8\}, \{9,10\}\big\}$$

  • These groups are counted separately: $$\big\{ \{\color{blue}1,\color{blue}2\}, \{3,4\}, \{5,6\}, \{7,8\}, \{9,10\}\big\}$$ $$\big\{ \{3,4\}, \{\color{blue}1,\color{blue}2\}, \{5,6\}, \{7,8\}, \{9,10\}\big\}$$

We account for unordered pairs, by diving by $2!$ for each group, since each pair has $2$ members. We also account for unordered groups, by dividing by $5!$, since there are $5$ pairs in each group. This gives $$\frac{10!}{2!^5 5!}=945.$$

The book answer uses a different counting approach.

  • Person $1$ is paired with someone, who can be one of $9$ people. After this pair is chosen there are $8$ remaining people.

  • After one pair is chosen, we identify a remaining person whose partner is one of $7$ remaining people, thus defining another pair.

  • After two pairs are chosen, we identify a remaining person whose partner is one of $5$ remaining people, thus defining another pair.

And so on. This gives $9 \times 7 \times 5 \times 3 \times 1=945$.

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Your evaluation counts as being different the three identical choices $(A,B), (C,D), ...$, $(B,A), (C,D),...$, and $(C,D), (A,B), ...$

List possibilities for 4 people and you will understand why the result is 3 and not $4!$.

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The permutation of 10 people is $10!$. Since the order of two people in a group is not important and the order of groups are not important,too, we divide 10! to 2!2!2!2!2!5!. Then, we have

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

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    $\begingroup$ The ordering of the groups themselves is not important either. $\endgroup$
    – Joce
    Mar 2, 2017 at 8:46

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