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
$=\,^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1}=13^{4}$
In an examination there are three multiple choice questions and each question has $4 $ choices. Number of ways in which a student can fail to get all answers correct, is
Total number of $6-$digit numbers in which only and all the five digits $1,3,5,7$ and $9$ appear, is
Let the number of elements in sets $A$ and $B$ be five and two respectively. Then the number of subsets of $A \times B$ each having at least $3$ and at most $6$ element is :
How many $6 -$ digit numbers can be formed from the digits, $0,1,3,5,7$ and $9$ which are divisible by $10$ and no digit is repeated?
If ${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}={ }^{\mathrm{q}} \mathrm{P}_{\mathrm{r}}-\mathrm{s}, 0 \leq \mathrm{s} \leq 1$ then ${ }^{\mathrm{q}+\mathrm{s}} \mathrm{C}_{\mathrm{r}-\mathrm{s}}$ is equal to .... .