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Suppose a class comprises eight girls and six boys. How many ways can a committee of three students of among 8 girls and 6 boys be formed if the committee must include at least one girl?

My solution:

Ways to form a committee with exactly $g$ girls:

$g = 1$: $\displaystyle\binom 8 1 \binom 6 2 = 120$

$g = 2$: $\displaystyle\binom 8 2 \binom 6 1 = 168$

$g = 3$: $\displaystyle\binom 8 3 = 56$

I have no idea what to do next.

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    $\begingroup$ at least one = all combination - combination without girls $\endgroup$ – lab bhattacharjee Jun 20 '13 at 17:46
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    $\begingroup$ You need to tell us how many boys and girls there are in the sample set... $\endgroup$ – apnorton Jun 20 '13 at 18:02
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I will assume from your work that there are $\bf 8$ girls AND $\bf 6$ boys (I will take your "9" in the second computation to be a typo), and you need to determine how many committees can be formed from those $14$ students that have at least one girl on the committee of three.

The "simpler approach" to this would be to take the total number of possible ways of selecting any committee of 3 students from among $ 8 + 6 = 14$ students, and then subtract from that total the number of committees/combinations that can be formed without girls. The result will give you the number of committees which then contain at least one girl.

$$\binom{14}{3} - \binom 63 = \text{total number of committees of 3 students with $\bf at\;least\; one\; girl$.}$$


Alternatively, you can add the results of your posted calculations, and arrive at the same number of valid combinations as you would using this approach. Both approaches give you a total of $344$ possible valid combinations.

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  • $\begingroup$ so the answer is 8? $\endgroup$ – user73122 Jun 20 '13 at 18:20
  • $\begingroup$ No recall that $\binom {14}{3} = 14\,C\,3 = \dfrac{14!}{11!\cdot 3!} = \dfrac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1} = \;?\;$, and $\binom 63 = \dfrac{6!}{3!\cdot 3!} = 20$. In all you should have $\;364 - 20 = 344$ $\endgroup$ – amWhy Jun 20 '13 at 18:24
  • $\begingroup$ Yes I understand that...from my giving solution what would be the next step.... I think that's the way the teacher wants it. $\endgroup$ – user73122 Jun 20 '13 at 18:26
  • $\begingroup$ Just add your totals up and see that you get = $344$! ;-) $\endgroup$ – amWhy Jun 20 '13 at 18:28
  • $\begingroup$ do you get the same answer using your approach? $\endgroup$ – user73122 Jun 20 '13 at 18:28
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girls boys product 8c1=8 6c2=15 120 8c2=28 6c1=6 168 8c3=56 6c0=1 +56 344

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  • $\begingroup$ Please add some words to give this some meaning. $\endgroup$ – dfeuer Nov 2 '13 at 13:28
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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. In particular, you may find helpful that $\binom{n}{k}$ gives $\binom{n}{k}$. $\endgroup$ – Cameron Buie Nov 2 '13 at 13:42

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