Question

If \( C_j = \binom{n}{j} \), then \( C_0 C_t + C_1 C_{t+1} + C_2 C_{t+2} + \dots + C_n = \)

• A. \( \frac{(2n)!}{(n - 2r)! (n + 2r)!} \)
• B. \( \frac{(2n)!}{(n - r)! (n + r)!} \)
• C. \( 2n C_t \)
• D. \( 2n C_{r+1} \)

Hint:

When dealing with sums of products of binomial coefficients, use factorial and binomial expansion properties to simplify.

Answer 1

The Correct Option is B

The given problem asks us to evaluate the sum: \[ C_0 C_t + C_1 C_{t+1} + C_2 C_{t+2} + \dots + C_n \] where \( C_j = \binom{n}{j} \). This sum represents a combination of binomial coefficients. To evaluate it, we need to use the property of binomial expansions. The sum of products of binomial coefficients can be rewritten in terms of factorials: \[ C_0 C_t + C_1 C_{t+1} + C_2 C_{t+2} + \dots + C_n = \frac{(2n)!}{(n - r)! (n + r)!} \] Hence, the correct answer is option (2).