What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
four cards are of the same suit,
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 four suits: diamond, club, spade, heart and there are $13$ cards of each suit. Therefore, there are $^{13} C _{4}$ ways of choosing $4$ diamonds. Similarly, there are $^{13} C _{4}$ ways of choosing $4$ clubs, $^{13} C _{4}$ ways of choosing $4$ spades and $^{13} C _{4}$ ways of choosing $4$ hearts. Therefore
The required number of ways $=\,^{13} C _{4}+^{13} C _{4}+^{13} C _{4}+^{13} C _{4}$
$=4 \times \frac{13 !}{4 ! 9 !}=2860$
The solution set of $^{10}{C_{x - 1}} > 2\;.{\;^{10}}{C_x}$ is
If $^{{n^2} - n}{C_2}{ = ^{{n^2} - n}}{C_{10}}$, then $n = $
The number of values of $'r'$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2}=^{69}C_{r^2-1} - ^{69}C_{3r}$ is :-
If $P(n,r) = 1680$ and $C(n,r) = 70$, then $69n + r! = $
Ten persons, amongst whom are $A, B$ and $C$ to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ and $B$ wants to speak before $C$ is