If a team of $4$ persons is to be selected out of $4$ married couples to play a mixed doubles tennis game,then the number of ways of forming a team in which no married couple appears is:

If ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C_7={ }^{14} C_{r}$,then the value of $r$ is

If $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$,then $r = \dots$

To fill $12$ vacancies,there are $25$ candidates,of which $5$ are from the scheduled caste. If $3$ of the vacancies are reserved for scheduled caste candidates while the rest are open to all,then the number of ways in which the selection can be made is:

If $N(n) = n \prod_{r=1}^{2023} (n^2 - r^2)$ for $n > 2023$,then find the value of ${}^{N}C_{N-1}$ when $n = 2024$.

If ${}^n C_r$ denotes the number of combinations of $n$ things taken $r$ at a time,then the expression ${}^n C_{r+1} + {}^n C_{r-1} + 2{}^n C_r$ equals