To prove the identity

$k \binom{kn}{n} = n \binom{k}{1} \binom{kn - k}{n - 1}$

using a combinatorial argument, we interpret both sides as counting the same thing in two different ways.

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Step 1: Problem Interpretation

Consider a group of $kn$ people, and we want to form a team of $n$ people. Each person belongs to one of $k$ distinct groups, each with $n$ members.

LHS: $k \binom{kn}{n}$

The left-hand side $k \binom{kn}{n}$ represents:

1. Selecting $n$ people from the total $kn$ people in all groups ($\binom{kn}{n}$).
2. Multiplying by $k$, accounting for the fact that one specific group (out of the $k$ groups) is chosen as the leader group.

RHS: $n \binom{k}{1} \binom{kn - k}{n - 1}$

The right-hand side represents:

1. Selecting a leader person from $n$ people within one of the $k$ groups ($n$ choices).
2. Selecting a specific group from the $k$ groups ($\binom{k}{1} = k$).
3. Selecting $n-1$ additional people from the remaining $kn - k$ people (excluding the group where the leader is chosen) ($\binom{kn-k}{n-1}$).

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Step 2: Equality Argument

Now, both sides count the same quantity—the total number of ways to:

1. Select a team of $n$ people from $kn$ people such that one person is marked as a "leader," and the leader's group is also marked as the "leader group."

Why they match:

• The LHS counts this scenario by directly picking $n$ people from all $kn$ and assigning one of the $k$ groups arbitrarily as the "leader group."
• The RHS counts this by:
  1. Selecting a leader from $n$ people in a specific group.
  2. Selecting that specific group as the "leader group."
  3. Choosing the remaining $n-1$ members of