$A$ $5$-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 $n$ 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:

$A$ 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:

The value of $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r} \cdot {}^{40}{C_{15}} \cdot {}^{20}{C_r} - {}^{35}{C_{15}} \cdot {}^{15}{C_r} \cdot {}^{25}{C_r}} \right)}$ is:

If $^n{P_4} = 30 \times {^n}{C_5}$,then $n = $

The digit in the unit place of the number $(183!) + (3^{183})$ is