Total number of $6-$digit numbers in which only and all the five digits $1,3,5,7$ and $9$ appear, is
$\frac{5}{2}(6 !)$
$5^6$
$\frac{1}{2}(6 !)$
$6!$
The number of ways of choosing $10$ objects out of $31$ objects of which $10$ are identical and the remaining $21$ are distinct, is
All the five digits numbers in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The $97^{th}$ number in the list does not contain the digit:
If $^n{P_r} = 840,{\,^n}{C_r} = 35,$ then $n$ is equal to
If $^{n} C _{9}=\,\,^{n} C _{8},$ find $^{n} C _{17}$