If $\alpha = ^mC_2$,then $^\alpha C_2$ is equal to

The number of four-letter words that can be formed using the letters of the word $BARRACK$ is

The remainder obtained when $1! + 2! + 3! + \dots + 11!$ is divided by $12$ is

There are three sections in a question paper,each section containing $4$ questions. If a candidate has to answer exactly $5$ questions from this paper such that at least one question is answered from each section,then the number of ways in which a candidate can make the choice of questions is

$^{80}C_{40}$ is not divisible by -

Let $m = (9n^2 + 54n + 80)(9n^2 + 45n + 54)(9n^2 + 36n + 35)$. The greatest positive integer which divides $m$ for all positive integers $n$ is: