What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
are face 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 $12$ face cards and $4$ are to be selected out of these $12$ cards. This can be done in $^{12} C _{4}$ ways.
Therefore, the required number of ways $=\frac{12 !}{4 ! 8 !}=495$
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 $N_1+N_2+N_3+N_4+N_5=$
The number of ways in which we can select three numbers from $1$ to $30$ so as to exclude every selection of all even numbers is
In an examination, a question paper consists of $12$ questions divided into two parts i.e., Part $\mathrm{I}$ and Part $\mathrm{II}$, containing $5$ and $7$ questions, respectively. A student is required to attempt $8$ questions in all, selecting at least $3$ from each part. In how many ways can a student select the questions?
If $a, b$ and $c$ are the greatest value of $^{19} \mathrm{C}_{\mathrm{p}},^{20} \mathrm{C}_{\mathrm{q}}$ and $^{21 }\mathrm{C}_{\mathrm{r}}$ respectively, then
If $^n{P_r}$=$ 720$.$^n{C_r},$ then $r$ is equal to