If ${ }^9 C_3+{ }^9 C_5={ }^{10} C_r$ for some $r \in N$,then $r=$

$A$ father with $8$ children takes $3$ children to the zoo at a time. If he can go more than once without taking the same $3$ children together,in how many ways can the father go to the zoo?

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:

The value of $\sum_{r=0}^4 {}^{(19-r)} C_3 + {}^{15} C_4$ is equal to

The number of groups that can be made from $5$ different green balls,$4$ different blue balls,and $3$ different red balls,if at least $1$ green and $1$ blue ball must be included.

Let $S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r, x, y, z, r \in \mathbb{Z}\}$ and $n(S_r)$ represents the number of elements in $S_r$. Then $n(S_2) + n(S_3) + n(S_4) = $