Number of all possible words (with or without meaning) that can be formed using all the letters of the word $CABINET$ in which neither the word $CAB$ nor the word $NET$ appear is

Words with or without meaning are to be formed using all the letters of the word $EXAMINATION$. The probability that the letter $M$ appears at the fourth position in any such word is:

Let the eleven letters $A, B, ....., K$ denote an arbitrary permutation of the integers $(1, 2, ....., 11)$. Then,what is the nature of the product $(A - 1)(B - 2)(C - 3) ..... (K - 11)$?

Consider all possible permutations of the letters of the word $ENDEANOEL$. Match the Statements / Expressions in $Column I$ with the Statements / Expressions in $Column II$.

$Column I$	$Column II$
$(A)$ The number of permutations containing the word $ENDEA$ is	$(p)$ $5!$
$(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is	$(q)$ $2 \times 5!$
$(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is	$(r)$ $7 \times 5!$
$(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is	$(s)$ $21 \times 5!$

Among the inequalities below,which ones are true for all natural numbers $n > 1000$?
$I. n! \leq n^n$
$II. (n!)^2 \leq n^n$
$III. 10^n \leq n!$
$IV. n^n \leq (2n)!$

Let $A$ be a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen, and the set $A$ is reconstructed by replacing the elements of $P$. $A$ subset $Q$ of $A$ is chosen again. The number of ways of choosing $P$ and $Q$ such that $Q$ contains just one element more than $P$ is