Combinatorics: 'At Least' Questions
Four representatives are to be randomly chosen from a committee of six men and six women. How many different four-representative combinations contain at least one woman?
Incorrect.
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Correct.
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First calculate the number of total arrangements. You need four
__SeBoxes__. Remember that the number decreases because there is no
repetition (you cannot choose the same person twice) and that the order
does
not matter because there aren't any different roles:
>$$\displaystyle \text{Total combinations} = \frac{\mathop{\fbox{12}}{{\times}}
\mathop{\fbox{11}}\times \mathop{\fbox{10}}\times \mathop{\fbox{9}}}
{4\times3\times2\times1}=11\times5\times9=495$$.
Now, calculate the number of forbidden options—those in which there are {color:dark-green}no
women, only __four men__{/color}. This is actually picking four people out of a limited selection of six men (again, not
ordered, because there are no roles):
>$$\displaystyle \text{Forbidden combinations} = \frac{\mathop{\fbox{6}}\times\mathop{\fbox{5}}{{\times}}
\mathop{\fbox{4}}\times \mathop{\fbox{3}}}
{4\times3\times2\times1}=5\times3=15 $$.
Lastly, subtract the number of "only men" quartets from the total number of
quartets to find the number of "at least one woman" quartets:
>$$\text{Good combinations} = 495 - 15 = 480$$.
Incorrect.
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Incorrect.
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Incorrect.
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400
415
455
480
495
