In how many ways can a committee of $5$ members be formed from $6$ men and $4$ women such that the committee contains at least one woman?

Compute $\frac{12!}{10! \times 2!}$

In how many ways can a committee of $6$ members be formed from $8$ men and $4$ women such that the committee contains at least $3$ women?

At an election,a voter may vote for any number of candidates not exceeding the number to be elected. If $4$ candidates are to be elected out of the $12$ contested in the election and a voter votes for at least one candidate,then the number of ways in which a voter can vote is:

$A$ committee of $3$ persons is to be constituted from a group of $2$ men and $3$ women. In how many ways can this be done? How many of these committees would consist of $1$ man and $2$ women?

From a group of $10$ men and $8$ women,the number of ways of forming a committee of $8$ members with not more than $5$ men and not less than $5$ women is