Let $$N=\binom{20}{7}-\binom{20}{8}+\binom{20}{9}-\binom{20}{10}+\binom{20}{11}-\binom{20}{12}+.......-\binom{20}{20}.$$
Then $N$ is divisible by
$\bf{Options::}\;\;:: (a)\; 4\;\;\;\;\;\; (b)\;\; 7\;\;\;\;\;\; (c)\;\; 9\;\;\;\;\;\; (d)\;\; None\; of\; these.$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+..........+\binom{n}{n}x^n.$$
Now put $x=-1\;,n=20\;,$ We get $$0=\binom{20}{0}-\binom{20}{1}+\binom{20}{2}+.......+\binom{20}{20}=S$$
So we get $$-N=-\binom{20}{7}+\binom{20}{8}-\binom{20}{9}+\binom{20}{10}-\binom{20}{11}+\binom{20}{12}+.......+\binom{20}{20}$$
So $$-N = S-\binom{20}{0}+\binom{20}{1}-\binom{20}{2}+\binom{20}{3}-\binom{20}{4}+\binom{20}{5}-\binom{20}{6}$$
So we get $$N=\binom{20}{0}-\binom{20}{1}+\binom{20}{2}-\binom{20}{3}+\binom{20}{4}-\binom{20}{5}+\binom{20}{6}$$
So we get $$N=1-20+\frac{20\cdot 19}{2}-\frac{20\cdot 19\cdot 18}{3\cdot 2}+\frac{20\cdot 19\cdot 18 \cdot 17 }{4\cdot 3\cdot 2}-\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{ 5\cdot 4\cdot 3\cdot 2}+\frac{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15}{ 6\cdot 5\cdot 4\cdot 3\cdot 2}$$
Now How can I proceed after that.
or If there is any short method then plz explain here.
Thanks
You began as I would have. If you have a calculator, you can just find the value of $N$ directly. If you don't have a calculator or you want to avoid such calculations, you can proceed using only modular arithmetic, which is easy enough to do without a calculator.
Each fraction in your last formula for $N$ is an integer, of course. You can strike out common factors from the numerators and denominators, then simplify the calculation by using only modular arithmetic. To check the divisibility by $4$, use modulo $4$.
$$\begin{align} N\ &\equiv (1) - (20) + (10\cdot 19) - (20\cdot 19\cdot 3) + (5\cdot 19\cdot 3\cdot 17) - (19\cdot 3\cdot 17\cdot 16) + (19\cdot 17\cdot 8\cdot 15)\\ &\equiv (1)-(0)+(2\cdot -1)-(0\cdot 19\cdot 3)+(1\cdot -1\cdot -1\cdot 1)-(19\cdot 3\cdot 17\cdot 0)+(19\cdot 17\cdot 0\cdot 15) \\ &\equiv 1-0+-2-0+1-0+0 \\ &\equiv 0 \pmod 4 \end{align}$$
We conclude that $N$ is divisible by $4$.
Redo that modulo $7$ and $9$ to get your final answer. I get that $N$ is also divisible by $7$ but not by $9$. It seems your final answer should be both choices $(a)$ and $(b)$. Can your test/homework handle that?