If $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,then $\sum_{r=0}^n \frac{r}{^nC_r}$ equals

If $n$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero,then $n$ is equal to:

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots \text{ up to } 51^{\text{th}} \text{ term}) + (1! - 2! + 3! - \dots \text{ up to } 51^{\text{th}} \text{ term})$ is equal to

$3$ boys $B_i, i = 1, 2, 3$ and $6$ girls $G_i, i = 1, 2, . . . , 6$ are to be seated in a row. The number of ways they can be seated so that $B_1, B_2$ are separated and $G_1, G_2$ are also separated is equal to:

Three letters are chosen at random from the letters of the word $VARIABLE$ and all possible three-letter words (with or without meaning) are formed with them. Then the probability of getting a three-letter word having a consonant as its middle letter is

If $n = ^mC_2$,then the value of $^nC_2$ is given by