How many numbers consisting of $5$ digits can be formed in which the digits $3, 4$ and $7$ are used only once and the digit $5$ is used twice?

All possible numbers are formed using the digits $1, 1, 2, 2, 2, 2, 3, 4, 4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is

How many $4$-digit numbers are there with no digit repeated?

Let $b_{1} b_{2} b_{3} b_{4}$ be a $4$-element permutation with $b_{i} \in \{1, 2, 3, \ldots, 100\}$ for $1 \leq i \leq 4$ and $b_{i} \neq b_{j}$ for $i \neq j$,such that either $b_{1}, b_{2}, b_{3}$ are consecutive integers or $b_{2}, b_{3}, b_{4}$ are consecutive integers. Find the number of such permutations.

Let $P$ be the set of seven-digit numbers with the sum of their digits equal to $11$. If the numbers in $P$ are formed by using the digits $1, 2,$ and $3$ only,then the number of elements in the set $P$ is:

Find $r$ if $^{5}P_{r} = 2 \times ^{6}P_{r-1}$.