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 $=\,^{13} C _{4}+^{13} C _{4}+^{13} C _{4}+^{13} C _{4}$
$=4 \times \frac{13 !}{4 ! 9 !}=2860$
A man $X$ has $7$ friends, $4$ of them are ladies and $3$ are men. His wife $Y$ also has $7$ friends, $3$ of them are ladies and $4$ are men. Assume $X$ and $Y$ have no comman friends. Then the total number of ways in which $X$ and $Y$ together can throw a party inviting $3$ ladies and $3$ men, so that $3$ friends of each of $X$ and $Y$ are in this party is :
If $P(n,r) = 1680$ and $C(n,r) = 70$, then $69n + r! = $
The set $S = \left\{ {1,2,3, \ldots ,12} \right\}$ is to be partitioned into three sets $A,\,B,\, C$ of equal size . Thus $A \cup B \cup C = S$ અને $A \cap B = B \cap C = C \cap A = \emptyset $ . The number of ways to partition $S$ is
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
The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals