If $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$,then $x+y+z=$

The students $S_{1}, S_{2}, \ldots, S_{10}$ are to be divided into $3$ groups $A, B$ and $C$ such that each group has at least one student and the group $C$ has at most $3$ students. Then the total number of possibilities of forming such groups is ........ .

If $\sum\limits_{r = 0}^{25} {\left( {^{50}C_r \cdot ^{50 - r}C_{25 - r}} \right) = K\left( {^{50}C_{25}} \right)}$,then $K$ is equal to

$A$ question paper is divided into two parts $A$ and $B$,and each part contains $5$ questions. The number of ways in which a candidate can answer $6$ questions,selecting at least two questions from each part,is:

Let $a, b, c \in \mathbb{N}$ and $a+b+c=5$. Let $L$ and $M$ be the least and greatest values of $2^a 3^b 5^c$ respectively. Then $M-L=$

The value of $\sum\limits_{r = 1}^{15} {{r^2} \left( \frac{^{15}C_r}{^{15}C_{r - 1}} \right)}$ is equal to