$A$ $5$-digit number $\overline{abcde}$,when multiplied by $9$,gives the $5$-digit number $\overline{edcba}$. The sum of the digits in the number is:

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

If a number is chosen at random from the four-digit numbers formed by using the digits $0, 1, 2, 3, 4, 6$ without repetition,then the probability that it is divisible by $4$ is

$A$ question paper contains $4$ questions,each having $4$ alternative answers. The number of ways that a candidate can answer one or more questions is

Three and four digit numbers are formed from the digits $1, 3, 5, 6, 8$. If $e_1$ is the number of three-digit even numbers with no digit repeated and $e_2$ is the number of four-digit even numbers with no digit repeated. Also,$O_1$ represents the number of three-digit odd numbers in which no digit is repeated and $O_2$ represents the number of four-digit odd numbers in which no digit is repeated. Then:

An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)