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 number of $4$ letter words (with or without meaning) that can be formed from the eleven letters of the word $'EXAMINATION'$ is
In how many ways can $5$ red and $4$ white balls be drawn from a bag containing $10$ red and $8$ white balls
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 $m, n$
In an election there are $8$ candidates, out of which $5$ are to be choosen. If a voter may vote for any number of candidates but not greater than the number to be choosen, then in how many ways can a voter vote
The number of ways of choosing $10$ objects out of $31$ objects of which $10$ are identical and the remaining $21$ are distinct, is