A car will hold $2$ in the front seat and $1$ in the rear seat. If among $6$ persons $2$ can drive, then the number of ways in which the car can be filled is

${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if:

$\sum \limits_{ k =0}^6{ }^{51- k } C _3$ is equal to

The least value of a natural number $n$ such that $\left(\frac{n-1}{5}\right)+\left(\frac{n-1}{6}\right) < \left(\frac{n}{7}\right)$, where $\left(\frac{n}{r}\right)=\frac{n !}{(n-r) ! r !}, i$

$n$ balls each of weight $w$ when weighted in pairs the sum of the weights of all the possible pairs is $120$ when they are weighed in triplets the sum of the weights comes out to be $480$ for all possible triplets, then $n$ is

If $^n{C_r} = 84,{\;^n}{C_{r - 1}} = 36$ and $^n{C_{r + 1}} = 126$, then $n$ equals