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 ?