Key Concepts and Formulas

This problem primarily relies on two fundamental identities from the Binomial Theorem:

1. Sum of Binomial Coefficients: The sum of all binomial coefficients for a given $n$ is $2^n$. $\sum_{k=0}^n {}^n{C_k} = {}^n{C_0} + {}^n{C_1} + \dots + {}^n{C_n} = 2^n$ This can be easily seen by setting $x=1$ in the binomial expansion of $(1+x)^n$.

2. Sum of Squares of Binomial Coefficients (Vandermonde's Identity for a special case): The sum of the squares of binomial coefficients is equal to the middle binomial coefficient of ${}^{2n}{C_n}$. $\sum_{k=0}^n ({}^n{C_k})^2 = ({}^n{C_0})^2 + ({}^n{C_1})^2 + \dots + ({}^n{C_n})^2 = {}^{2n}{C_n}$ This identity can be derived by considering the coefficient of $x^n$ in the expansion of $(1+x)^n (x+1)^n = (1+x)^{2n}$.

Problem Statement We need to evaluate the sum: $S = \sum\limits_{\substack{i,j = 0 \\ i \ne j}}^n {{}^n{C_i}\,{}^n{C_j}}$

Step-by-Step Solution

1. Understanding the Summation and Choosing a Strategy The summation $S$ asks us to sum the product of two binomial coefficients, ${}^n{C_i}$ and ${}^n{C_j}$, for all possible pairs of indices $(i, j)$ from $0$ to $n$, with the crucial condition that $i \ne j$. Directly handling the condition $i \ne j$ can be cumbersome. A common and effective strategy for such problems is to consider the "total" sum (without any restrictions on $i$ and $j$) and then subtract the "unwanted" terms (where $i=j$).

So, we can write the given sum as: $S = \left( \sum\limits_{i=0}^n \sum\limits_{j=0}^n {{}^n{C_i}\,{}^n{C_j}} \right) - \left( \sum\limits_{i=j=0}^n {{}^n{C_i}\,{}^n{C_j}} \right)$ Let's break this down into two parts:

• Part A: The total sum $T = \sum\limits_{i=0}^n \sum\limits_{j=0}^n {{}^n{C_i}\,{}^n{C_j}}$ (where $i$ and $j$ can be equal).
• Part B: The sum of terms where $i=j$, denoted as $E = \sum\limits_{i=j=0}^n {{}^n{C_i}\,{}^n{C_j}}$.

Then, the required sum $S = T - E$.

2. Evaluating Part A: The Total Sum ($T$) The total sum is given by: $T = \sum\limits_{i=0}^n \sum\limits_{j=0}^n {{}^n{C_i}\,{}^n{C_j}}$ Explanation: Since the terms ${}^n{C_i}$ and ${}^n{C_j}$ are independent with respect to the indices $i$ and $j$ in a double summation, we can separate the double summation into a product of two single summations. $T = \left( \sum\limits_{i=0}^n {{}^n{C_i}} \right) \left( \sum\limits_{j=0}^n {{}^n{C_j}} \right)$ Explanation: Now, we apply the first key identity: $\sum_{k=0}^n {}^n{C_k} = 2^n$. Both parentheses contain this standard sum. $T = (2^n) \cdot (2^n)$ $T = 2^{n+n}$ $T = 2^{2n}$

3. Evaluating Part B: The Sum where $i=j$ ($E$) The sum of terms where $i=j$ is given by: $E = \sum\limits_{i=j=0}^n {{}^n{C_i}\,{}^n{C_j}}$ Explanation: Since $i=j$, we can replace $j$ with $i$ in the expression, simplifying it to a single summation. $E = \sum\limits_{i=0}^n {{}^n{C_i}\,{}^n{C_i}}$ $E = \sum\limits_{i=0}^n ({}^n{C_i})^2$ Explanation: Now, we apply the second key identity: $\sum_{k=0}^n ({}^n{C_k})^2 = {}^{2n}{C_n}$. $E = {}^{2n}{C_n}$

4. Combining the Results Finally, we substitute the values of $T$ and $E$ back into our initial decomposition for $S$: $S = T - E$ $S = 2^{2n} - {}^{2n}{C_n}$

This matches option (A).

Tips and Common Mistakes

• Understanding Summation Limits: Always pay close attention to the indices and their ranges. Here, $i$ and $j$ both range from $0$ to $n$.
• Condition $i \ne j$: This is a classic indicator that you might need to use the "total minus unwanted" strategy. Don't try to enumerate pairs directly.
• Binomial Identities are Your Friends: Memorizing and understanding the derivations of common binomial identities like $\sum {}^n{C_k} = 2^n$ and $\sum ({}^n{C_k})^2 = {}^{2n}{C_n}$ is crucial for solving problems involving binomial coefficients efficiently.
• Factorization of Double Sums: Remember that $\sum_i \sum_j f(i)g(j) = (\sum_i f(i))(\sum_j g(j))$. A common mistake is to incorrectly try to factorize when the terms are not independent (e.g., if it were $\sum_i \sum_j f(i,j)$).

Summary and Key Takeaway

This problem demonstrates an elegant application of fundamental binomial identities. The key takeaway is that sums with restrictive conditions like $i \ne j$ can often be simplified by considering the complete sum and then subtracting the cases that violate the condition. Mastering the identities $\sum_{k=0}^n {}^n{C_k} = 2^n$ and $\sum_{k=0}^n ({}^n{C_k})^2 = {}^{2n}{C_n}$ is essential for success in combinatorics problems.