Let's walk through the problems one by one, proving the combinatorial identities stated. All of them are related to binomial coefficients and properties derived from them.

Problem Setup:

Given that $n \geq k \geq r \geq 0$ and $k > 0$, we need to prove the following combinatorial identities.

The binomial coefficient $C(n, k)$ represents the number of ways to choose $k$ elements from $n$ elements, and it is defined as: $C(n, k) = \frac{n!}{k!(n-k)!}$

Part (a)

Identity to Prove: $C(n, k) C(k, r) = C(n, r) C(n - r, k - r)$ Proof: This identity can be derived by considering the combinatorial interpretation. The left-hand side counts the number of ways to:

1. Choose $k$ elements from $n$, and then
2. Choose $r$ elements from those $k$ elements.

The right-hand side counts the number of ways to:

1. Choose $r$ elements from $n$, and then
2. Choose $k - r$ elements from the remaining $n - r$ elements.

Thus, both sides count the same quantity, and the identity holds.

Part (b)

Identity to Prove: $C(n, k) C(k, r) = C(n, k - r) C(n - k + r, r)$ Proof: This can be understood by first recognizing that on the left-hand side, we choose $k$ elements from $n$, and then choose $r$ elements from those $k$. On the right-hand side:

• We choose $k - r$ elements from $n$, and then we choose $r$ elements from the remaining $n - k + r$.

This symmetry in how we can split the selections proves the equality.

Part (c)

Identity to Prove: $\sum_{j=0}^n C(n, j) C(j, r) = C(n, r) 2^{n-r}$ Proof: Consider the left-hand side as summing over all ways to choose $j$ elements from $n$, and then choosing $r$ elements from those $j$. This represents selecting $r$ elements from a subset of $n$, while the factor $2^{n-r}$ accounts for all possible ways to fill in the remaining $n - r$ elements with either inclusion or exclusion. Hence, the identity holds.

Part (d)

Identity to Prove: $\sum_{j=k}^n (-1)^{j+k} C(n, j) = C(n - 1, k - 1)$ Proof: This is a combinatorial identity that can be derived from the principle of inclusion-exclusion. The left-hand side sums over all ways to choose $j$ elements from $n$, with alternating signs, while the right-hand side represents choosing $k - 1$ elements from $n - 1$. The alternating signs come from inclusion-exclusion reasoning, and this sums up to the value given on the right-hand side.

Let me know if you'd like further details on any step or clarification.

Questions for further practice:

1. Can you prove a similar identity for sums involving binomial coefficients, like $\sum_{j=0}^n C(n, j) = 2^n$?
2. What is the combinatorial meaning of $C(n, k)$ in the context of Pascal’s triangle?
3. How can you derive the Vandermonde identity from these binomial relations?
4. How does inclusion-exclusion apply to combinatorial proofs of binomial identities?
5. Can you think of another combinatorial identity where alternating signs appear in summation?

Tip: Many binomial identities can be proved either through algebraic manipulation (using factorials) or by using combinatorial arguments (counting different ways to select elements). Both methods are valuable and can provide deep insight into how the identities work.