An urn contains $5$ red marbles,$4$ black marbles and $3$ white marbles. The number of ways in which $4$ marbles can be drawn so that at most three of them are red is

$A$ committee of $4$ persons is to be formed from $2$ ladies,$2$ old men,and $4$ young men such that it includes at least $1$ lady,at least $1$ old man,and at most $2$ young men. The total number of ways in which this committee can be formed is:

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

$A$ committee of $5$ is to be formed out of $6$ men and $4$ ladies. The number of ways this can be done,when at most $2$ ladies are included,is

$A$ set contains $(2n + 1)$ elements. The number of subsets of the set which contain at most $n$ elements is:

At an election,a voter may vote for any number of candidates,not greater than the number to be elected. There are $10$ candidates and $4$ are to be selected. If a voter votes for at least one candidate,then the number of ways in which he can vote is: