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 $(7!)$)

The total number of three-digit and five-digit integers which can be formed by using the digits $0, 1, 2, 3, 4, 5$,using each digit not more than once in each number,is:

Let $a_1, a_2, \ldots, a_n$ be $n$ non-zero real numbers,of which $p$ are positive and the remaining are negative. The number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is positive is $55$. Similarly,the number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is negative is $50$. Then,the value of $p^2 + (n-p)^2$ is

The total number of natural numbers of six digits that can be made with digits $1, 2, 3, 4$,if all digits are to appear in the same number at least once,is

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

There are $3$ applicants for a scholarship in Physics,$5$ for Mathematics,and $4$ for Chemistry. In how many different ways can one of these scholarships be awarded?