How many numbers greater than $1000000$ can be formed by using the digits $1,2,0,2,4,2,4 ?$
since, $1000000$ is a $7 -$ digit number and the number of digits to be used is also $7$. Therefore, the numbers to be counted will be $7 -$ digit only. Also, the numbers have to be greater than $1000000$ , so they can begin either with $1,2$ or $ 4.$
The number of numbers beginning with $1=\frac{6 !}{3 ! 2 !}=\frac{4 \times 5 \times 6}{2}=60,$ as when $1$ is fixed at the extreme left position, the remaining digits to be rearranged will be $0,2,2,2, $$4,4,$ in which there are $3,2 s$ and $2,4 s$
Total numbers begining with $2$
$=\frac{6 !}{2 ! 2 !}=\frac{3 \times 4 \times 5 \times 6}{2}=180$
and total numbers begining with $4=\frac{6 !}{3 !}=4 \times 5 \times 6=120$
Therefore, the required number of numbers $=60+180+120=360$
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?
There are $5$ students in class $10,6$ students in class $11$ and $8$ students in class $12.$ If the number of ways, in which $10$ students can be selected from them so as to include at least $2$ students from each class and at most $5$ students from the total $11$ students of class $10$ and $11$ is $100 \mathrm{k}$, then $\mathrm{k}$ is equal to $......$
$^n{P_r}{ \div ^n}{C_r}$ =
The total number of ways of selecting six coins out of $20$ one rupee coins, $10$ fifty paise coins and $7$ twenty five paise coins is
The number of ordered pairs ( $\mathrm{r}, \mathrm{k}$ ) for which $6 \cdot ^{35} \mathrm{C}_{\mathrm{r}}=\left(\mathrm{k}^{2}-3\right)\cdot{^{36} \mathrm{C}_{\mathrm{r}+1}}$. where $\mathrm{k}$ is an integer, is