Combinatorics: 'At Least' Questions
How many five-person committees chosen at random from a group consisting of five men, five women, and five children contain at least one woman?
Correct.
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For the total possible committees, you're picking five people from fifteen (five women, five men, and five children) with no order and no repetition:
>$$\displaystyle C(15{,}5)=\frac{15!}{10!\times5!}=
\frac{15\times14\times13\times12\times11}{5\times4\times3
\times2\times1}=7\times3\times13\times11$$.
Now, for the forbidden options (committees with no women at all), you are picking five persons out of ten (just the men and the children):
>$$\displaystyle C(10{,}5)=\frac{10!}{5!\times5!}=
\frac{10\times9\times8\times7\times6}{5\times4\times3
\times2\times1}=3\times2\times7\times6$$.
Thus, the number of __good combinations__ is
>$$\text{Good combinations} = 7\times 3\times 13\times 11 - 3\times 2\times 7\times 6$$.
Wait—before you go and figure out each number separately, extract the common factors $$3\times 7$$:
>$$\text{Good combinations} = 3\times 7\times (13\times 11 - 2\times 6)$$
>>>>>$$=21\times (143-12)$$
>>>>>$$=21\times 131$$
>>>>>$$=(20+1)\times 131$$
>>>>>$$=2{,}751$$
Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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700
1,221
1,434
2,751
3,011
