Proving the Equivalence of Factorial Expressions in Combinatorics

In combinatorics, proving the equivalence of factorial expressions is a fundamental concept. One such example is the equation (frac{n1!}{(k-1!(n-k)!)} frac{n1!}{(k!(n-k_1)!)).

Understanding the Problem

The given equation can be simplified to verify its validity. Let's break down the components and provide a detailed explanation.

Left-Hand Side

Starting with the left-hand side of the equation:

(frac{n1!}{(k-1!(n-k)!)).

Here, (n1!) can be rewritten using the properties of factorials. Specifically, we know that:

(n1! n1 cdot n!).

Substituting this into the left-hand side, we get:

(frac{n1 cdot n!}{(k-1!(n-k)!)).

We can further simplify this by expressing (k-1!) as:

(k-1! frac{k!}{k}).

Therefore, the left-hand side can be rewritten as:

(frac{n1 cdot n!}{frac{k!}{k} cdot (n-k)!} frac{n1 cdot n! cdot k}{k! cdot (n-k)!}).

Right-Hand Side

Now let's consider the right-hand side of the equation, which is:

(frac{n1!}{k!(n-k_1)!).

Again, using the fact that (n1! n1 cdot n!), we can rewrite the right-hand side as:

(frac{n1 cdot n!}{k!(n-k_1)!}).

Comparing Both Sides

Notice that the (n1 cdot n!) term is common in both sides. We need to compare their denominators:

Left-hand side: (frac{k}{k!} cdot (n-k)! frac{k cdot (n-k)!}{k!}).

Right-hand side: (k!(n-k_1)!).

By simplifying the right-hand side, we can see that:

(k!(n-k_1)! k! cdot (n-k)!)

Therefore, we can rewrite the left-hand side as:

(frac{n1 cdot n! cdot k}{k! cdot (n-k)!} frac{n1 cdot n!}{k!(n-k)!}).

Comparing both sides, we see that they are indeed equal:

(frac{n1!}{(k-1!(n-k)!)} frac{n1!}{(k!(n-k_1)!)})

Conclusion

The equivalence of these formal expressions is proven. This result is significant in combinatorics as it provides a method to count the number of ways to choose (k) objects from a set of (n1) objects. This can be done directly or by removing one object and counting the subsets of (k) and (k-1) objects.

Combinatorial Interpretation

On the right-hand side, the expression (frac{n1!}{k!(n-k_1)!}) counts the number of ways to choose (k) objects from a set of (n1) objects.

This can be understood by considering two cases:

Removing one object to form a subset of (k-1) objects and then adding it back to form a subset of (k) objects, or Choosing (k) objects directly from the initial set.

Both methods yield the same result, confirming the equivalence of the expressions.