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 $=60+180+120=360$
Team $'A'$ consists of $7$ boys and $n$ girls and Team $'B'$ has $4$ boys and $6$ girls. If a total of $52$ single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then $n$ is equal to
If $n \geq 2$ is a positive integer, then the sum of the series ${ }^{ n +1} C _{2}+2\left({ }^{2} C _{2}+{ }^{3} C _{2}+{ }^{4} C _{2}+\ldots+{ }^{ n } C _{2}\right)$ is ...... .
The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to
The least value of a natural number $n$ such that $\left(\frac{n-1}{5}\right)+\left(\frac{n-1}{6}\right) < \left(\frac{n}{7}\right)$, where $\left(\frac{n}{r}\right)=\frac{n !}{(n-r) ! r !}, i$