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JEE Main 202226 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2022 · 26 July · Evening Shift · Mathematics · Binomial Theorem

∑ _ i, j = 0 _ i ≠ j ^ n Ci n Cj n is equal to

∑i,j=0i≠jnCinCjn\mathop{∑}_{\mathop{i, j = 0}_{i \ne j}}^{n} Ci n Cj n is equal to

Options

  1. A22n−Cn2n{2}^{2n} - Cn 2n
  2. B22n−1−Cn−12n−1{2}^{2n - 1} - Cn - 1 2n - 1
  3. C22n−12Cn2n{2}^{2n} - \dfrac{1}{2} Cn 2n
  4. D2n−1+Cn2n−1{2}^{n - 1} + Cn 2n - 1

Correct answer

A. 22n−Cn2n{2}^{2n} - Cn 2n

Step-by-step solution

∑i,j=0i≠jnCinCjn\mathop{∑}_{\mathop{i, j = 0}_{i \ne j}}^{n} Ci n Cj n

=C0n(C1n+C2n+....Cnn)+C1n(C0n+C2n+....Cnn)+....Cnn(C0n+C1n+....Cn−1n)= C0 n \left(C1 n + C2 n + . . . . Cn n\right) + C1 n \left(C0 n + C2 n + . . . . Cn n\right) + . . . . Cn n \left(C0 n + C1 n + . . . . Cn - 1 n\right)

=2(C0nC1n+C1nC2n+....Cn−1nCnn)= 2\left(C0 n C1 n + C1 n C2 n + . . . . Cn - 1 n Cn n\right)

=(C0n+C1n....Cnn)2−[(C0n)2+(C1n)2.....(Cnn)2]= {\left(C0 n + C1 n . . . . Cn n\right)}^{2} - \left[{\left(C0 n\right)}^{2} + {\left(C1 n\right)}^{2} . . . . . {\left(Cn n\right)}^{2}\right]

=(2n)2−Cn2n=22n−Cn2n= {\left({2}^{n}\right)}^{2} - Cn 2n = {2}^{2n} - Cn 2n

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