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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?

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2 Answers 2

up vote 5 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 $\binom{12+10}{5}-\binom{12}{1}\binom{10}{4}-\binom{12}{0}\binom{10}{5}$.

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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

$$\binom{12}2\binom{10}3+\binom{12}3\binom{10}2+\binom{12}4\binom{10}1+\binom{12}5\;,$$

but it’s pretty clearly more efficient to calculate

$$\binom{22}5-\binom{12}1\binom{10}4-\binom{10}5\;.$$

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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
2  
@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

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