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By how many ways can a committee of 6 people formed out of 8 men and 4 women if there are particular two men refuse to be together ?

My attempt: The total number of choosing the committee is

$C^8_6 + C^8_5 \times C^4_1 + C^8_4 \times C^4_2 + C^8_3 \times C^4_3 + C^8_2 \times C^4_4$

The number of ways to choose the two men who refuse to be together is

$C^8_2 \times (C^4_1+C^4_2 + C^4_3 + C^4_4) $ So we can find the required by subtracting the last result from the first result

Is my answer correct ?

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Your general approach is correct, but the second part is wrong.

For the total number without considering the condition, a simpler formula is $\binom{8+4}{6}=924$, which is equal to your $\sum_{k=0}^4 \binom{4}{k}\binom{8}{6-k}$ via Vandermonde's identity.

For determining what to subtract from this, the two particular men are already specified, so you don't choose them. Instead, you choose the remaining $6-2$ committee members from the remaining $12-2$ people, yielding $$924 - \binom{10}{4} = 924 - 210 = 714.$$

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