# Combinatorics problem involving teams (Discrete math)

A company has 10 men and 18 women. A work team consists of two workers. What is the maximum number of work teams (man-man,woman-woman,man-woman) that can be formed from this group? How many different ways can this maximum number of work teams be formed? I know the max amount of teams is 14. I tried 28 choose 2 for the maximum number of team combinations but it's wrong.

• Could you clarify how the gender of the workers enters into the problem? Nov 24, 2015 at 22:23
• A team can consist of either a man and a woman, or two of one or the other. Nov 24, 2015 at 22:24
• Are you asking how many ways the $28$ workers can be paired off into $14$ teams of two people? Nov 24, 2015 at 22:26

Imagine pairing people off one group at a time. For the first group, you've got $28$ choose $2$ possible ways to pick a team of $2$ from $28$ people. After you've selected that team, you've got to pick another team of $2$ from the remaining $26$ people, so $26$ choose $2$. Continue on in that manner until you've assigned all the teams, and the total number of possibilities so far is: $${28 \choose 2} \cdot {26 \choose 2} \cdots {2 \choose 2} = \frac{28!}{2! \cdot 26!} \cdot \frac{26!}{2!\cdot 24!} \cdots \frac{2!}{2! \cdot 0!} = \frac{28!}{2^{14}}.$$

This gives you all the different ways you could pick teams if the order you were picking them mattered (i.e., if you were picking a "team 1", "team 2", etc.). Since it doesn't, you want to divide this result by the number of labellings you can give to the same set of $14$ teams, which is $14!$. Hence, the number of ways of splitting 28 people into teams of two is $${28 \choose 2} \cdot {26 \choose 2} \cdots {2 \choose 2} \cdot \frac{1}{14!} = \frac{28! \cdot 26! \cdots 2!}{2^{14} \cdot 26! \cdot 24! \cdots 2! \cdot 14!} = \frac{28!}{2^{14} \cdot 14!}$$

Since gender doesn't restrict anything, and people are unique individuals, we just need to count ways to divide twenty-eight people into fourteen pairs.

There are $28!$ ways to line everyone up, then split into $14$ pairs.   However, each pair can be formed in $2!$ ways, and we don't care about the $14!$ ways to arrange the teams either.

$$\dfrac{28!}{{2!}^{14}\,14!}$$

Graham Kemp and user164385 have provided you with nice solutions. Here is an alternative approach.

Line up the $28$ workers in some order, whether alphabetically, by seniority, or some other criterion. The first person in line has $27$ ways of selecting a work partner. The next person in line who is not already on a team has $25$ ways of selecting a work partner. Continue. The number of teams that can be selected is $$27 \cdot 25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = \prod_{k = 1}^{14} (2k - 1) = 27!!$$ Observe that if we multiply the numerator and denominator of this answer by $2^{14}14!$, we obtain $$\frac{28!}{2^{14}14!}$$