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. Then the total number of ways in which this committee can be formed is
$40$
$41$
$16$
$32$
Let $\left(\begin{array}{l}n \\ k\end{array}\right)$ denotes ${ }^{n} C_{k}$ and $\left[\begin{array}{l} n \\ k \end{array}\right]=\left\{\begin{array}{cc}\left(\begin{array}{c} n \\ k \end{array}\right), & \text { if } 0 \leq k \leq n \\ 0, & \text { otherwise }\end{array}\right.$
If $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$
and $A_{4}-A_{3}=190 \mathrm{p}$, then $p$ is equal to :
Value of $r$ for which $^{15}{C_{r + 3}} = {\,^{15}}{C_{2r - 6}}$ is
The number of seven digit positive integers formed using the digits $1,2,3$ and $4$ only and sum of the digits equal to $12$ is $...........$.
$^n{C_r}{ + ^n}{C_{r - 1}}$ is equal to
An urn contains $5$ red marbles, $4$ black marbles and $3$ white marbles. Then the number of ways in which $4$ marbles can be drawn so that at the most three of them are red is