# The number of ways to form a committee with a condition

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 ?

## 1 Answer

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 $$924 - \binom{10}{4} = 924 - 210 = 714.$$