There are 8 boys and 10 girls needed to form 5 boy-girl teams (2 people per team). How many different teams are possible?
My understanding is that $${\binom {8} 5}\times \binom {10} 5$$
would be the number of ways to select a group of ten people (5 boys and 5 girls). But to actually seperate into 5 teams of two, boy and girl in each, we would need to select each team
First team: $8\times 10$ choices,
Second team: $7 \times 9$ choices,
…
Fifth team: $4 \times 6 $ choices
Then finally, divide through by the number of permutations of 5 teams so
The answer should be:
$$
\frac{(8\times 10) \cdot (7\times 9) \cdot (6\times 8) \cdot (5\times 7) \cdot (4\times 6) }{5!} = \frac{^{10}P_5\cdot {^8}P_5}{5!}=1,693, 440
$$
Is this correct or is my logic flawed somewhere?
Best Answer
Your reasoning is sound, and your answer is correct. You could also have argued as follows, starting with your first observation. There are, as you say, $\binom85\binom{10}5$ ways to choose $5$ boys and $5$ girls. Now line up the boys. There are $5$ ways to pick a teammate for the first boy in line, $4$ ways to pick a teammate for the second boy in line, and so on, so there are $5!$ ways to match up the boys and the girls. That gives you a total of $\binom85\binom{10}55!=1\,693\,440$ different ways to form $5$ teams, each consisting of a boy and a girl.