Consider choosing $k$ items from a set of $n$. This is equivalent to choosing the $n-k$ items that you don't want to include. For example, if you have 5 fruits and want to choose 2, there are $\binom{5}{2} = 10$ ways. This is the same as choosing the 3 fruits you don't want, $\binom{5}{3} = 10$. This property is extremely useful for simplifying calculations.

This identity is visually represented in Pascal's Triangle, where each number is the sum of the two numbers directly above it. For instance, to find $\binom{5}{2}$, you add $\binom{4}{1}$ and $\binom{4}{2}$. This property is crucial for recursive algorithms and understanding combinatorial identities.

Consider the binomial expansion $(x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n}y^n$. If we set $x=1$ and $y=1$, we get $(1+1)^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}$. This simplifies to $2^n = \sum_{k=0}^{n} \binom{n}{k}$. This property is useful for counting problems where all possible subsets are considered.

This property can be derived from the binomial expansion of $(1-1)^n$. For $n > 0$, $(1-1)^n = 0$. Expanding this gives $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} = 0$. This identity is useful in problems involving alternating sums or when dealing with inclusion-exclusion principles.