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
$=\,^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1}=13^{4}$
The number of ways in which thirty five apples can be distributed among $3$ boys so that each can have any number of apples, is
${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if:
The value of ${}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} $ is
Let $\left(\begin{array}{l}n \\ k\end{array}\right)$ denotes ${ }^{n} C_{k}$ and $\left[\begin{array}{l} n \\ k \end{array}\right]=\left\{\begin{array}{cc}\left(\begin{array}{c} n \\ k \end{array}\right), & \text { if } 0 \leq k \leq n \\ 0, & \text { otherwise }\end{array}\right.$
If $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$
and $A_{4}-A_{3}=190 \mathrm{p}$, then $p$ is equal to :
Six ‘$+$’ and four ‘$-$’ signs are to placed in a straight line so that no two ‘$-$’ signs come together, then the total number of ways are