Dividing ten people into five two-person groups. 
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?
 A: 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 $9 \times 7 \times 5 \times 3 \times 1=945$.
A: 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 $4!$.
A: 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
$\frac{10!}{2!2!2!2!2!5!}$
