prove that (n choose k) = (n-1 choose k-1) (n-1 choose k)

a) By using formula of (n choose k)
b) by breaking subsets of size k into two mutually exclusive classes, one class comprising all those subsets which contain a given element and the other all those which dont

To prove the equation (n choose k) = (n-1 choose k-1) (n-1 choose k), we can use two different approaches:

a) By using the formula of (n choose k):

The formula for (n choose k), also known as binomial coefficient, is given by:

(n choose k) = n! / (k! * (n-k)!)

Using this formula, let's evaluate the right-hand side of the equation:

(n-1 choose k-1) (n-1 choose k) = (n-1)! / ((k-1)! * (n-1-k+1)!) * (n-1)! / (k! * (n-1-k)!)

Simplifying this expression, we get:

(n-1 choose k-1) (n-1 choose k) = (n-1)! / ((k-1)! * (n-k)!) * (n-1)! / (k! * (n-1-k)!)

Now, notice that (n-1)! / ((k-1)! * (n-k)!) is the same as (n choose k) since we are just replacing n with (n-1) and k with (k-1).

Similarly, (n-1)! / (k! * (n-1-k)!) is also the same as (n choose k) as we are replacing n with (n-1) and k with k.

Therefore, we can rewrite the right-hand side of the equation as (n choose k) * (n choose k), which is equal to (n choose k) squared.

So, we have (n-1 choose k-1) (n-1 choose k) = (n choose k)^2

b) By breaking subsets of size k into two mutually exclusive classes:

Consider a set of n elements, and we want to choose k elements from this set.

Let's focus on one particular element, say element X. Any subset of size k either contains X or doesn't contain X.

1. Subsets containing X:
For a subset of size k to contain X, we need to choose (k-1) elements from the remaining (n-1) elements. So, there are (n-1 choose k-1) ways to choose subsets that contain X.

2. Subsets not containing X:
For a subset of size k to not contain X, we need to choose k elements from the remaining (n-1) elements (excluding X). So, there are (n-1 choose k) ways to choose subsets that don't contain X.

Since these two classes of subsets (containing X and not containing X) are mutually exclusive, we can add the counts together to get the total number of subsets of size k from a set of n elements.

Therefore, (n choose k) = (n-1 choose k-1) + (n-1 choose k).

By substituting this back into the original equation, we have (n choose k) = (n-1 choose k-1) (n-1 choose k), which proves the given equation.