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).

Answer 2

Given:
We are given the identity involving binomial coefficients:
\[ \sum_{j = 0}^{n} \binom{n}{j} \binom{n}{t + j} \]
This is a known identity from combinatorics involving sums of products of binomial coefficients.

Standard Identity Used:
There exists a standard identity for sums of this form:
\[ \sum_{j = 0}^{n} \binom{n}{j} \binom{n}{r + j} = \binom{2n}{n + r} \]
Here, the indices are shifted such that:
- \( C_j = \binom{n}{j} \)
- \( C_{t + j} = \binom{n}{t + j} \)
- So we are summing over \( \binom{n}{j} \binom{n}{t + j} \), from \( j = 0 \) to \( n \)

This directly matches the identity above, with \( r = t \).

Therefore:
\[ \sum_{j = 0}^{n} \binom{n}{j} \binom{n}{t + j} = \binom{2n}{n + t} \]
Using factorial notation:
\[ \binom{2n}{n + t} = \frac{(2n)!}{(n + t)! (n - t)!} \]

Final Answer:
\[ \boxed{\frac{(2n)!}{(n - t)! (n + t)!}} \]