What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
four cards belong to four different suits,
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 $13$ cards in each suit.
Therefore, there are $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of diamond, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of hearts, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of clubs, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of spades. Hence, by multiplication principle, the required number of ways
$=\,^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1}=13^{4}$
Number of different words that can be formed from all letters of word $APPLICATION$ such that two vowels never come together is -
If $n$ and $r$ are two positive integers such that $n \ge r,$ then $^n{C_{r - 1}}$$ + {\,^n}{C_r} = $
In how many ways a team of $11$ players can be formed out of $25$ players, if $6$ out of them are always to be included and $5$ are always to be excluded
From a class of $25$ students, $10$ are to be chosen for an excursion party. There are $3$ students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Find the number of ways of selecting $9$ balls from $6$ red balls, $5$ white balls and $5$ blue balls if each selection consists of $3$ balls of each colour.