# A class of children contains 14 boys and 12 girls. How many ways are there to choose a team of six children from the class if…

A class of children contains 14 boys and 12 girls. How many ways are there to choose a team of six children from the class if it must contain at least two boys and at least two girls?

I already know the answer(191191), which was found by considering the cases of 2 boys + 4 girls, 3 boys + 3 girls and, 4 boys and 2 girls separately - then adding up the numbers.

However I am not sure why I can't do the following, so would like the error to be pointed out:

First choose 2 boys, then choose two girls, then choose any two pupils from the remaining 22. So

2 boys = 14C2 combinations = 91
2 girls = 12C2 combinations = 66
2 remaining pupils = 22C2 combinations = 231

Total number = 91*66*231 = 1387386

Your problem is that you count some teams several times.

The same set (Albert, Bill, Cindy, Daisy, Evan, Flint) will be counted 6 times by your method:

• (Albert, Bill) + (Cindy, Daisy) + (Evan, Flint)
• (Albert, Evan) + (Cindy, Daisy) + (Bill, Flint)
• (Bill, Evan) + (Cindy, Daisy) + (Albert, Flint)

etc.

This is a frequent problem in combinatorics: You are overcounting (which you can see by your number that is too high) because the same choice is counted several times.

Because, call the boys $B_1,...B_{14}$ and the girls $G_1,...,G_{12}$, then in your way of counting, you could say:

I take $B_1$ and $B_2$, then $G_1$ and $G_2$. Now two arbitrary pupils, let's say $B_5$ and $G_7$.

But you could also start with the boys $B_2$ and $B_5$, then take the girls $G_1$ and $G_7$ and for the arbitrary pupils, take $B_1$ and $G_2$. Now you have the same overall team, but you counted as though they were different possibilities.

It is very difficult to correct such an overcounting, so it's better to start counting the right way.

You may understand what's wrong if you think of the following example. Say you want to choose 2 boys out of 4.

Then the correct way is $\binom{4}{2}=6$. But if you do it the way you are suggesting it will be $\binom{4}{1}\binom{3}{1}=12$. In other words, you take into account the order, hence you have more ways to choose them.

Try this

Conditions: atleast 2 boys and atleast 2 girls and a team of 6 to be formed

Boys-2 ,girls -4 C(14,2) . C(12,4)=91 .495=45045

Boys-3, girls -3 C(14,3).C(12,3)=104 . 220= 80080

Boys-4,girls-2 C(14,4).C(12,2)=1001 . 66=66066