What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
two are red cards and two are black cards,
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 $=\,^{52} C _{4}=\frac{52 !}{4 ! 48 !}=\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}$
$=270725$
There are $26$ red cards and $26$ black cards. Therefore, the required number of ways $=^{26} C _{2} \times^{26} C _{2}$
$=\left(\frac{26 !}{2 ! 24 !}\right)^{2}=(325)^{2}=105625$
A person wants to climb a $n-$ step staircase using one step or two steps. Let $C_n$ denotes the number of ways of climbing the $n-$ step staircase. Then $C_{18} + C_{19}$ equals
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
If $^{{n^2} - n}{C_2}{ = ^{{n^2} - n}}{C_{10}}$, then $n = $
The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets themselves such that they get consecutive blocks of $5, 3$ and $2$ tickets is
Let $S=\{1,2,3, \ldots ., 9\}$. For $k=1,2, \ldots \ldots, 5$, let $N_K$ be the number of subsets of $S$, each containing five elements out of which exactly $k$ are odd. Then $N_1+N_2+N_3+N_4+N_5=$