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