JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q4

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 4 of 90

$\displaystyle\sum_{\substack{i,j=0\\i\ne j}}^{n}{}^nC_i\,{}^nC_j$ is equal to

2^{2n}-{}^{2n}C_n

2^{2n-1}-{}^{2n-1}C_{n-1}

2^{2n}-\dfrac12\,{}^{2n}C_n

2^{n-1}+{}^{2n-1}C_n

Answer and worked solution

Official answer: A

All pairs minus equal pairs. $\sum_{i,j}{}^nC_i{}^nC_j=\left(2^n\right)^2$ , and the terms with $i=j$ add up to $\sum\left({}^nC_i\right)^2={}^{2n}C_n$ .

So the sum is $2^{2n}-{}^{2n}C_n$ .

(A)

Topics:Properties of binomial coefficients

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 4