Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We must choose a 5-member team from 12 girls and 10 boys. How many ways are there to make the choice so that there are no more than 3 boys on the team?

The correct answer is $\binom{22}{5} - \binom{12}{1} \binom{10}{4} - \binom{10}{5}$.

I understand the $\binom{22}{5}$ part, but where I am confused at is the other two parts. I do not know how to get those parts. Can anyone help me understand how to get to the solution?

share|cite|improve this question
up vote 6 down vote accepted

From the total number of ways to form a $5$ member team we subtract the numbers corresponding to the cases when exactly $4$ boys or exactly $5$ boys are chosen to form teams that have no more than $3$ boys. The first case happens when we choose $4$ boys and $1$ girl and the second happens when we choose $5$ boys and $0$ girls. This gives us $\binom{12+10}{5}-\binom{12}{1}\binom{10}{4}-\binom{12}{0}\binom{10}{5}$.

share|cite|improve this answer
Thank you. I didn't think it would be that simple. – Wooooop Dec 9 '12 at 7:53

That solution proceeds by starting with $\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\binom{12}1\binom{10}4$ teams that have one girl and four boys and the $\binom{10}5$ teams that have five boys.

The problem could also be solved by noting that there are $\binom{12}2\binom{10}3$ teams with two girls and three boys, $\binom{12}3\binom{10}2$ teams with three girls and two boys, $\binom{12}4\binom{10}1$ teams with four girls and one boy, and $\binom{12}5$ teams with five girls, and calculating the sum


but it’s pretty clearly more efficient to calculate


share|cite|improve this answer
I haven't thought about doing it that way! Brian, you have been helping me a lot, and I really appreciate that someone with your knowledge is on here helping others out. Thank you. – Wooooop Dec 9 '12 at 7:57
@kevlar: I enjoy it, and you’re very welcome. Quite a few problems of this general type can be attacked either by subtracting bad cases or adding good ones; it’s worth thinking of both possibilities and then deciding which one will involve less calculation. – Brian M. Scott Dec 9 '12 at 8:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.