To prove the given identity combinatorially:

Identity to prove:

$\binom{n}{r} \binom{r}{k} = \binom{n}{k} \binom{n-k}{r-k}$

Explanation:

Step 1: Interpret $\binom{n}{r} \binom{r}{k}$

• $\binom{n}{r}$: This counts the number of ways to choose $r$ elements from a set of $n$ elements.
• $\binom{r}{k}$: After choosing the $r$ elements, this counts the number of ways to select $k$ elements from the $r$ chosen elements.
• Thus, $\binom{n}{r} \binom{r}{k}$ counts the total number of ways to:
  1. First choose $r$ elements from $n$, and then
  2. Choose $k$ elements from the $r$.

Step 2: Interpret $\binom{n}{k} \binom{n-k}{r-k}$

• $\binom{n}{k}$: This counts the number of ways to choose $k$ elements from $n$ elements.
• $\binom{n-k}{r-k}$: After choosing $k$ elements, this counts the number of ways to select the remaining $r-k$ elements from the $n-k$ elements not yet chosen.
• Thus, $\binom{n}{k} \binom{n-k}{r-k}$ counts the total number of ways to:
  1. First choose $k$ elements from $n$, and then
  2. Choose $r-k$ elements from the remaining $n-k$ elements.

Step 3: Equivalence

Both sides count the same thing: the number of ways to select:

1. A total of $r$ elements from a set of $n$, and
2. Among those $r$, designate $k$ elements as a special subset.

This proves the identity combinatorially, as both sides represent the same counting process but in a different order.

Would you like further clarification or details on this proof?

Related questions to explore:

1. What is the difference between algebraic and combinatorial proofs for such identities?
2. How can this identity be derived algebraically?
3. How does this relate to Pascal’s triangle or other combinatorial relationships?
4. Can this identity be applied to real-world problems in statistics or probability?
5. How would you generalize this identity for other similar expressions?

Tip:

Always try to interpret binomial coefficients as counting methods; this helps simplify and validate combinatorial proofs effectively!