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

The correct answer is (A) : \(2^{2n \text\_2n}C_n\)
\(\sum_{\substack{i,j=0 \\ t \neq j}}^n\) \(^nC_i\ ^nC_j \)
\(= ∑^{n}_{i,j = 0}\) \(^nC_i\ ^nC_j - ∑^{n}_{i=j}\ ^nC_i\ ^nC_j\)
\(= ∑^{n}_{j=0}\ ^nC_i ∑^{n}_{j =0}\ ^nC_j - ∑^{n}_{ i =0}\ ^nC_i\ Ci\)
\(= 2^n.2^n-\ ^{2n}C_n\)
\(= 2^{2n\_2n}C_n\)