Question

\(\sum_{\substack{i,j=0 \\ t \neq j}}^n\) \(^nC_i\ ^nC_j \)
is equal to

• A. \(2^{2n \text\_2n}C_n\)
• B. \(2^{2n-1\_2n-1}C_{n-1}\)
• C. \(2^{2n-\frac{1}{2}}\ ^{2n}C_n\)
• D. \(2^{n-1}+2^{2n-1}C_n\)

Answer 1

The Correct Option is A

To find the value of the expression \(\sum_{\substack{i,j=0 \\ t \neq j}}^n\) \(^nC_i\ ^nC_j\), let's analyze it step-by-step.

The given expression can be interpreted as the sum of products of binomial coefficients, where sums are taken over all possible indices \(i\) and \(j\) except where \(i = j\).

To simplify this, we start with the identity involving binomial expansions: \((1+x)^n = \sum_{k=0}^{n} {n \choose k} x^k\)

If we expand two separate binomial identities and multiply them: \((1+x)^n (1+x)^n = \left(\sum_{i=0}^{n} {n \choose i} x^i\right) \left(\sum_{j=0}^{n} {n \choose j} x^j\right)\)This gives: \((1+x)^{2n} = \sum_{i=0}^{n} \sum_{j=0}^{n} {n \choose i} {n \choose j} x^{i+j}\)

The term \(x^t\) in this expansion is given by: \(\sum_{i+j=t} {n \choose i} {n \choose j}\)The sum over all terms except where \(i = j\) can be written as \(\sum_{i=0}^{2n} \left( \sum_{i+j=t} {n \choose i} {n \choose j} \right) - \sum_{i=0}^{n} \left( {n \choose i} {n \choose i} \right)\)

For the full expansion \((1+1)^{2n}\), the total is: \(2^{2n}\)But we subtract the diagonal \(\sum_{i=0}^{n} {n \choose i}^2\), which equals: \(\sum_{k=0}^{n} {n \choose k}^2 = {2n \choose n}\)(using the identity: \(\sum_{k=0}^{n} {n \choose k}^2 = {2n \choose n}\)).

Therefore, the value of \(\sum_{\substack{i,j=0 \\ t \neq j}}^n\ ^nC_i\ ^nC_j\) is: \(2^{2n} - {2n \choose n} = 2^{2n} - {2n \choose n}\)

The correct option provided matches this analysis, which is \(2^{2n} - {2n \choose n}\).