Question

$\left({}^{30}{C}_0\right)\left({}^{30}{C}_{10}\right)-\left({}^{30}{C}_1\right)\left({}^{30}{C}_{11}\right)+.....\left({}^{30}{C}_{20}\right)\left({}^{30}{C}_{30}\right)$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is

A

Each of the terms has two coefficients.
We can re-write the terms as follows
$\left({}^{30}{C}_0\right)\left({}^{30}{C}_{20}\right)-{}^{30}{C}_1\left({}^{30}{C}_{19}\right)....+{}^{30}{C}_{20}{}^{30}{C}_0$
If we look at the terms now, the sum of bottom numbers in the coefficients add up to 20.
$(e.g.{}^{30}{C}_0{}^{30}{C}_{20}$; here 0 + 20 = 20)
At this point we should be able to guess that each term is formed by multiplying two terms from two binomial expansions. One of them should have negative terms also.
Let us consider the expansion of
$(1+x{)}^{30}(1-x{)}^{30}$,
$(1+x{)}^{30}(1-x{)}^{30}$
$=({}^{30}{C}_0+{}^{30}{C}_{1}x+{}^{30}{C}_2{x}^2.....{}^{30}{C}_{10}{x}^{10}+.....{}^{30}{C}_{30}{x}^{30})\times ({}^{30}{C}_0-{}^{30}{C}_{1}x+{}^{30}{C}_2{x}^2.....{}^{30}{C}_{10}{x}_{10}+.....{}^{30}{C}_{30}{x}^{30})$
We will find the coefficient of $x^{20}$
= ${}^{30}{C}_0{}^{30}{C}_{20}+{}^{30}{C}_1\times \left({}^{-30}{C}_{19}\right)+{}^{30}{C}_2{}^{30}{C}_{18}.....\left({}^{30}{C}_{20}\right)\left({}^{30}{C}_0\right)$
This is the required sum.
We can say that the required sum is the coefficient $x^{20}$ in the expansion of $(1+x{)}^{30}(1-x{)}^{30}or(1-x^2{)}^{30}$.
Coefficient of $x^{20}$ in the expansion of $(1-x^2{)}^{30}={}^{30}{C}_{10}$
$[T_{r+1}={}^{30}{C}_r\times (-x^2{)}^r$
$\Rightarrow$ 2r = 20
$\Rightarrow$ r = 10
$\Rightarrow$ coefficient = ${}^{30}{C}_{10}$