How many $6 -$ digit numbers can be formed from the digits, $0,1,3,5,7$ and $9$ which are divisible by $10$ and no digit is repeated?

$\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)$, whenever $0 \le r \le n - 1$is equal to

Let $\left(\begin{array}{l}n \\ k\end{array}\right)$ denotes ${ }^{n} C_{k}$ and $\left[\begin{array}{l} n \\ k \end{array}\right]=\left\{\begin{array}{cc}\left(\begin{array}{c} n \\ k \end{array}\right), & \text { if } 0 \leq k \leq n \\ 0, & \text { otherwise }\end{array}\right.$

If $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$

and $A_{4}-A_{3}=190 \mathrm{p}$, then $p$ is equal to :

A group of $9$ students, $s 1, s 2, \ldots, s 9$, is to be divided to form three teams $X, Y$ and, $Z$ of sizes $2,3$ , and $4$, respectively. Suppose that $s_1$ cannot be selected for the team $X$, and $s_2$ cannot be selected for the team $Y$. Then the number of ways to form such teams, is. . . .

$6$ different letters of an alphabet are given. Words with four letters are formed from these given letters. Then the number of words which have atleast one letter repeated and no two same letters are together, is

For $2 \le r \le n,\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) + 2\,\left( \begin{array}{l}\,\,n\\r - 1\end{array} \right)$ $ + \left( {\begin{array}{*{20}{c}}n\\{r - 2}\end{array}} \right)$ is equal to