$3$ boys $B_i, i = 1, 2, 3$ and $6$ girls $G_i, i = 1, 2, . . . , 6$ are to be seated in a row. The number of ways they can be seated so that $B_1, B_2$ are separated and $G_1, G_2$ are also separated is equal to:

$A$ linguistic club consists of $6$ girls and $4$ boys. $A$ team of $4$ members is to be selected from this group including the selection of a leader (from among these $4$ members) for the team. If the team has to include at most one boy,the number of ways of selecting the team is

$A$ letter lock consists of three rings with $15$ different letters each. If $N$ denotes the number of ways in which it is possible to make unsuccessful attempts to open the lock, then:

Consider all possible permutations of the letters of the word $ENDEANOEL$. Match the Statements / Expressions in $Column I$ with the Statements / Expressions in $Column II$.

$Column I$	$Column II$
$(A)$ The number of permutations containing the word $ENDEA$ is	$(p)$ $5!$
$(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is	$(q)$ $2 \times 5!$
$(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is	$(r)$ $7 \times 5!$
$(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is	$(s)$ $21 \times 5!$

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

Find the number of natural number solutions to the equation $x_1 + x_2 = 100$,such that $x_1$ and $x_2$ are not multiples of $5$.