The number of positive integral solutions of the equation $xyz = 90$ is equal to :-

If all the letters of the word $ACADEMICIAN$ are permuted in all possible ways,then the number of permutations in which no two $A$'s are together and all the consonants are together is:

Five boys and five girls are writing an examination in a hall in which $5$ benches are arranged in a row and only two students are to be seated on every bench at either of its ends. If the seating arrangement is to be such that no two boys or no two girls sit together as neighbours (a student should not have a student of same gender either on left or right),then the total number of such arrangements 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:

For all $n \in N$,the product $(n+24)(n+25)(n+26)(n+27)$ is always divisible by:

Suppose $a_2, a_3, a_4, a_5, a_6, a_7$ are integers such that $\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$,where $0 \leq a_j < j$ for $j = 2, 3, 4, 5, 6, 7$. The sum $a_2 + a_3 + a_4 + a_5 + a_6 + a_7$ is: