The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:

Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^3$ in $(1+x)^3+(1+x)^4+(1+x)^5+\ldots+(1+x)^{99}+(1+k x)^{100}, x \neq 0$, is $\left(43 n+\frac{101}{4}\right)\left({ }^{100} \mathrm{C}_3\right)$ for some $n \in \mathrm{~N}$, be $p$. Then the value of $p+n$ is :

Suppose that the mean and median of the non-negative numbers $21,8,17, a, 51,103, b, 13,67,(a>b)$, are 40 and 21 , respectively. If the mean deviation about the median is 26 , then $2 a$ is equal to :

Let the line $\mathrm{L}_1: x+3=0$ intersect the lines $\mathrm{L}_2: x-y=0$ and $\mathrm{L}_3: 3 x+y=0$ at the points A and B , respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point $C$. Then $B C^2: A C^2$ is equal to: