Let $x$ denote the number of ways of selecting at least one ball from a bag containing $3$ identical red balls,$4$ identical blue balls,and $5$ identical green balls. Let $y$ denote the number of ways in which a student can fail in an examination,when they have to write the examination in $5$ different subjects. Then $x+y=$

Let $n \geq 3$ be an integer. For a permutation $\sigma = (a_1, a_2, \ldots, a_n)$ of $(1, 2, \ldots, n)$,we define $f_\sigma(x) = a_n x^{n-1} + a_{n-1} x^{n-2} + \ldots + a_2 x + a_1$. Let $S_\sigma$ be the sum of the roots of $f_\sigma(x) = 0$ and let $S$ denote the sum over all permutations $\sigma$ of $(1, 2, \ldots, n)$ of the values $S_\sigma$. Then,

Find the remainder when $(1!)^2 + (2!)^2 + (3!)^2 + \dots + (100!)^2$ is divided by $10^2$.

The total number of positive integral solutions $(x, y, z)$ of $xyz = 24$ is

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:

If all the letters of the word $COMBINATION$ are arranged in all possible ways to form $11$ letter words (with or without meaning),then the number of words among them in which $C$ and $N$ occupy the end positions and no vowel appears exactly in the middle position is