What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these
two are red cards and two are black 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 $26$ red cards and $26$ black cards. Therefore, the required number of ways $=^{26} C _{2} \times^{26} C _{2}$
$=\left(\frac{26 !}{2 ! 24 !}\right)^{2}=(325)^{2}=105625$
In an election the number of candidates is $1$ greater than the persons to be elected. If a voter can vote in $254$ ways, then the number of candidates is
If $n$ and $r$ are two positive integers such that $n \ge r,$ then $^n{C_{r - 1}}$$ + {\,^n}{C_r} = $
A committee of $7$ has to be formed from $9$ boys and $4$ girls. In how many ways can this be done when the committee consists of:
at most $3$ girls?
An engineer is required to visit a factory for exactly four days during the first $15$ days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1$-15$ June $2021$ is. . . . . .
The number of ways, in which $5$ girls and $7$ boys can be seated at a round table so that no two girls sit together, is