JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
$\sum_{\substack{i,j=0\\ i\ne j}}^{n}{}^nC_i\,{}^nC_j$ is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. The sum over all pairs $(i,j)$ is a product of two sums; remove the pairs with $i=j$. Formulas used. (1) $\sum_{i=0}^{n}{}^{n}C_i=2^n$(2) $\sum_{i=0}^{n}\left({}^{n}C_i\right)^2={}^{2n}C_n$Step 1 - All pairs. By (1), $$\sum_{i=0}^{n}\sum_{j=0}^{n}{}^{n}C_i\,{}^{n}C_j=\left(2^n\right)^2=2^{2n}.$$ Step 2 - Pairs with $i=j$. By (2), they add up to ${}^{2n}C_n$. Step 3 - Subtract. $\displaystyle\sum_{i\ne j}{}^{n}C_i\,{}^{n}C_j=2^{2n}-{}^{2n}C_n$. Answer: (A)