There are $10$ points in a row. In how many ways can $4$ points be selected such that no two of them are consecutive?

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$,then $n=$

In a lottery,a person chooses six different natural numbers at random from $1$ to $20$. If these six numbers match the six numbers already fixed by the lottery committee,he wins the prize. What is the probability of winning the prize in the game? [Hint: The order of the numbers is not important.]

$A$ bag contains $23$ balls, of which $7$ are identical. The number of ways of selecting $12$ balls from the bag is:

Consider three boxes,each containing $10$ balls labelled $1, 2, \dots, 10$. Suppose one ball is randomly drawn from each of the boxes. Denote by $n_i$ the label of the ball drawn from the $i^{th}$ box,$(i = 1, 2, 3)$. Then,the number of ways in which the balls can be chosen such that $n_1 < n_2 < n_3$ is:

If $^{n + 1}C_3 = 2 \cdot ^nC_2$,then $n =$