∑i,j=0i≠jnCinCjn is equal to

Correct answer: (A) 2^2n-Cn2n. ∑i,j=0i≠jnCinCjn =C0nC1n+C2n+....Cnn+C1nC0n+C2n+....Cnn+....CnnC0n+C1n+....Cn-1n =2C0nC1n+C1nC2n+....Cn-1nCnn =C0n+C1n....Cnn^2-C0n^2+C1n^2.....Cnn^2 =2n^2-Cn2n=2^2n-Cn2n

Q34 / 90 · JEE Main 2022 (26 Jul Shift 2) · Mathematics

$\sum_{\underset{i \neq j}{i, j = 0}}^{n} {}^{n}C_{i} {}^{n}C_{j}$ is equal to

Correct answer
A
$2^{2 n} - {}^{2 n}C_{n}$
B
$2^{2 n - 1} - {}^{2 n - 1}C_{n - 1}$
C
$2^{2 n} - \frac{1}{2} {}^{2 n}C_{n}$
D
$2^{n - 1} + {}^{2 n - 1}C_{n}$

Solution

$\sum_{\underset{i \neq j}{i, j = 0}}^{n} {}^{n}C_{i} {}^{n}C_{j}$
$= {}^{n}C_{0} ({}^{n}C_{1} + {}^{n}C_{2} + . . . . {}^{n}C_{n}) + {}^{n}C_{1} ({}^{n}C_{0} + {}^{n}C_{2} + . . . . {}^{n}C_{n}) + . . . . {}^{n}C_{n} ({}^{n}C_{0} + {}^{n}C_{1} + . . . . {}^{n}C_{n - 1})$
$= 2 ({}^{n}C_{0} {}^{n}C_{1} + {}^{n}C_{1} {}^{n}C_{2} + . . . . {}^{n}C_{n - 1} {}^{n}C_{n})$
$= {({}^{n}C_{0} + {}^{n}C_{1} . . . . {}^{n}C_{n})}^{2} - [{({}^{n}C_{0})}^{2} + {({}^{n}C_{1})}^{2} . . . . . {({}^{n}C_{n})}^{2}]$
$= {(2^{n})}^{2} - {}^{2 n}C_{n} = 2^{2 n} - {}^{2 n}C_{n}$

Hence, the correct answer is option: 'A'.

∑i,j=0i≠jnCinCjn is equal to

Solution

Relevant Questions

$\sum_{\underset{i \neq j}{i, j = 0}}^{n} {}^{n}C_{i} {}^{n}C_{j}$ is equal to