The number of ways in which a committee of $6$ members can be formed from $8$ gentlemen and $4$ ladies so that the committee contains at least $3$ ladies is:

Let $n(A) = 3$ and $n(B) = 3$ (where $n(S)$ denotes the number of elements in set $S$). Then,the number of subsets of $(A \times B)$ having an odd number of elements is:

There are $6$ Men and $8$ Women in a club in which a committee of $5$ people has to be made. What will be the number of ways to select $5$ people if at most one man is selected?

The number of four-digit natural numbers which contain exactly two distinct digits is:

The value of $r$ for which $^{20}C_r \cdot ^{20}C_0 + ^{20}C_{r-1} \cdot ^{20}C_1 + ^{20}C_{r-2} \cdot ^{20}C_2 + \dots + ^{20}C_0 \cdot ^{20}C_r$ is maximum 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: