What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
cards are of the same colour?
There will be as many ways of choosing $4$ cards from $52$ cards as there are combinations of $52$ different things, taken $4$ at a time. Therefore
The required number of ways $=\,^{52} C _{4}=\frac{52 !}{4 ! 48 !}=\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}$
$=270725$
$4$ red cards can be selected out of $26$ red cards in $^{26} C _{4}$ ways.
$4$ black cards can be selected out of $26$ black cards in $^{26} C _{4}$ ways.
Therefore, the required number of ways $=\,^{26} C _{4}+^{26} C _{4}$
$=2 \times \frac{26 !}{4 ! 22 !}=29900$
The value of ${}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} $ is
If all the six digit numbers $x_1 x_2 x_3 x_4 x_5 x_6$ with $0 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6$ are arranged in the increasing order, then the sum of the digits in the $72^{\text {th }}$ number is $............$.
For non-negative integers $s$ and $r$, let
$\binom{s}{r}=\left\{\begin{array}{ll}\frac{s!}{r!(s-r)!} & \text { if } r \leq s \\ 0 & \text { if } r>s\end{array}\right.$
For positive integers $m$ and $n$, let
$(m, n) \sum_{ p =0}^{ m + n } \frac{ f ( m , n , p )}{\binom{ n + p }{ p }}$
where for any nonnegative integer $p$,
$f(m, n, p)=\sum_{i=0}^{ p }\binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}$
Then which of the following statements is/are $TRUE$?
$(A)$ $(m, n)=g(n, m)$ for all positive integers $m, n$
$(B)$ $(m, n+1)=g(m+1, n)$ for all positive integers $m, n$
$(C)$ $(2 m, 2 n)=2 g(m, n)$ for all positive integers $m, n$
$(D)$ $(2 m, 2 n)=(g(m, n))^2$ for all positive integers $m, n$
If $^{n}{P_4} = 24.{\,^n}{C_5},$ then the value of $n$ is
The least value of a natural number $n$ such that $\left(\frac{n-1}{5}\right)+\left(\frac{n-1}{6}\right) < \left(\frac{n}{7}\right)$, where $\left(\frac{n}{r}\right)=\frac{n !}{(n-r) ! r !}, i$