# probability of N same type pairs from a heterogeneous group

There are 16 people: 8 males and 8 females. The people are randomly divided to pairs, each pair get the exact same mission. Let $$X$$ be the number of male-male pairs.

1. calculate $$P\{X=3\}$$
2. calculate $$P\{X=i\}$$ where $$i$$ is each value $$X$$ can take
3. find $$E[X]$$

for (1) there are $$\binom{8}{2,2,2,1,1}$$ ways to divide the males to 3 all male pairs and $$\binom{8}{2,2,2,1,1}$$ ways to divide the females to 3 all female pairs and there are 4 ways to pair the remaining males and females so I expected the probability to be $$P\{X=3\}=\frac{\binom{8}{2,2,2,1,1}\cdot \binom{8}{2,2,2,1,1} \cdot 4}{\binom{16}{2,2,2,2,2,2,2,2}} = \frac{\frac{8!}{8}\cdot \frac{8!}{8}\cdot 4}{\frac{16!}{2^8}}=\frac{8!^2\cdot 4\cdot 64 \cdot 4}{16! \cdot 64}=\frac{8!^2}{15!} \approx 0.0012$$ but it is incorrect so instead I tried, I choose 6 males and pair them in $$\binom{8}{6}\binom{6}{2,2,2}$$ and the 2 remaining males I have $$\binom{2}{1}\binom{8}{1}$$ and $$\binom{7}{1}$$ ways to pair with a female, and the remaining 6 females I have $$\binom{6}{2,2,2}$$ ways to pair so $$\frac{\binom{8}{6}\binom{6}{2,2,2}\binom{2}{1}\binom{8}{1}\binom{7}{1}\binom{6}{2,2,2}}{\binom{16}{2,2,2,2,2,2,2,2}}=\frac{\frac{8!}{16}\cdot 16 \cdot 7\cdot\frac{6!}{8}}{\frac{16!}{2^8}}=\frac{7!\cdot 7! \cdot 2^8}{16!}\approx 0.0003$$ which is also incorrect. I'm not sure how I can calculate $$P\{X=3\}$$ or any other $$P\{X=i\}$$

after reading @saulspatz's answer I found a possible solution: all possible pairing is given by $$\binom{16}{2,2,2,2,2,2,2,2}\cdot \frac{1}{8!}$$ because all the pairs have the same role but the same works for each of the 3 male-male or female-female pairs, hence $$\binom{6}{2,2,2}\cdot \frac{1}{3!}$$ and because the 2 male-female pairs also have the same role I get $$P\{X=3\}=\frac{\frac{\binom{8}{6}\binom{6}{2,2,2}}{3!}\frac{\binom{2}{1}\binom{8}{1}\binom{7}{1}}{2!}\frac{\binom{6}{2,2,2}}{3!}}{\frac{\binom{16}{2,2,2,2,2,2,2,2}}{8!}}=\frac{\frac{8!}{16\cdot 3!}\cdot \frac{16\cdot 7}{2!}\cdot\frac{6!}{8\cdot 3!}}{\frac{16!}{2^8 \cdot 8!}}=\frac{8!\cdot 7! \cdot 7!\cdot 2^8}{16!\cdot 3!\cdot 3!\cdot 2!}\approx 0.1740$$ and from here calculating $$P\{X=i\}$$ is the same and and there is no problem finding $$E[X]$$

• What is a mission and what role does it play in your question (if any)? Also, you say some results you calculated are incorrect. How do you know they are incorrect? (I am not saying they are correct, I am just pointing out that you made unsupported claims in your post.) – mathguy Nov 28 '20 at 16:34
• @mathguy I think the mission part is so that the order of the pairs doesn't matter. I have 4 possible answers and I need to choose one – CforLinux Nov 28 '20 at 16:48

The number of ways to divide the $$16$$ people is not pairs is not $$\binom{16}{2,2,2,2,2,2,2,2}$$ but $$\frac1{8!}\binom{16}{2,2,2,2,2,2,2,2}$$ To divide the people into pairs, arrange them into a line, and pair the first and second, the third and fourth, and so on. The order of the tow people in the pairs doesn't matter, so we have $$\frac{16!}{(2!)^8}$$ but also the order of the pairs themselves doesn't matter, so we must divide by $$8!$$.