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 $0 \leq r \leq n$. If ${ }^{n+1} C_{r+1} : { }^{n} C_{r} : { }^{n-1} C_{r-1} = 55 : 35 : 21$,then $2n + 5r$ is equal to:

Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5=20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is

Let $S$ be the set of all permutations $a_1, a_2, \ldots, a_6$ of $1, 2, \ldots, 6$ such that $a_1, a_2, \ldots, a_k$ is not a permutation of $1, 2, \ldots, k$ for any $k, 1 \leq k \leq 5$. Then the number of elements in $S$ is:

The number of different words that can be formed using all the letters of the word $APPLICATION$ such that no two vowels come together is -