Medium MCQ +4 / -1 PYQ · JEE Mains 2022

$$\sum\limits_{\matrix{ {i,j = 0} \cr {i \ne j} \cr } }^n {{}^n{C_i}\,{}^n{C_j}} $$ is equal to

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

Solution

<p>$$\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } } $$</p> <p>$$ = \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } } $$</p> <p>$= {2^n}\,.\,{2^n} - {}^{2n}{C_n}$</p> <p>$= {2^{2n}} - {}^{2n}{C_n}$</p>

About this question

Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion

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