Five-digit numbers are formed using the digits $1, 2, 3, 4, 5, 6$,and $8$. What is the probability that they have even digits at both the ends?

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

All five-letter words are made using all the letters $A, B, C, D, E$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_{n}$. Let the probability $P(W_{n})$ of choosing the word $W_{n}$ satisfy $P(W_{n}) = 2P(W_{n-1}), n > 1$. If $P(CDBEA) = \frac{2^{\alpha}}{2^{\beta}-1}, \alpha, \beta \in N$,then $\alpha + \beta$ is equal to

The number of five-digit numbers that are divisible by $6$ which can be formed by choosing digits from $\{0, 1, 2, 3, 4, 5\}$,when repetition is allowed,is

$A$ five-digit number divisible by $3$ is to be formed using the digits $0, 1, 2, 3, 4, 5$ without repetition. The total number of ways this can be done is:

If $n$ is an integer with $0 \leq n \leq 11$,then the minimum value of $n!(11-n)!$ is attained when a value of $n$ equals to