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a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.

case 1: neither are included -> 5C3 X 6C2

case 2: only youngest woman included -> 5C3 X 7C4

case 3: only youngest man included -> 6C3 X 6C3

i then add all of those up together. is this right?

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marked as duplicate by lulu, N. F. Taussig combinatorics Nov 27 '18 at 12:17

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  • $\begingroup$ All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman. $\endgroup$ – lulu Nov 27 '18 at 11:37
  • $\begingroup$ This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant. $\endgroup$ – lulu Nov 27 '18 at 12:01
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Without the age restriction, there would be $\binom{6}{3}\binom{7}{4}$. Of these, exactly $\binom{5}{2}\binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $\binom{6}{3}(\binom{7}{4}-\binom{5}{2})=20\times (35-10)=500$.

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This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.

In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.

Case 3 looks correct to me.

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