Binomial day

We know that

$\large\ \left( 1+x \right) ^{ 30 } = {30 \choose 0} + {30 \choose 1}x ^ 1 + {30 \choose 2}x ^ 2....$

$\large\ \left( x-1 \right) ^{ 30 } = {30 \choose 10}x ^ {20} - {30 \choose 11}x ^ {19} + {30 \choose 12}x ^ {18} - ....$ .

Multiplying respective terms on RHS , We get

$\large\ \left( x ^ 2 - 1 \right)^{ 30 } = \left[{30 \choose 0}{30 \choose 10} - {30 \choose 1}{30 \choose 11} + ... \right]x ^ {20}$ .

So, its enough to calculate coefficient of $\large\ x ^ {20}$ in the expansion of $\left( x ^ 2 - 1 \right)^{ 30 }$ , which is

$\boxed{\displaystyle{\large\ {30 \choose 10}}}$ .

Rewrite the summand as $(-1)^r$ times the product of 30Cr and 30C(20-r).Now, it is clear that the summation is nothing but the coefficient of x^20 in [(1-x)^30][(1+x)^30].This is same as coefficient of x^20 in (1-x^2)^30 which is equal to 30C10.Hence,the answer is 10.(Note that nCr is notation for n choose r).