Sum of Binomial Coefficients Equals Power of Two

Prove that the sum of binomial coefficients for a set of size $n$ equals $2^n$, i.e., sum_{k=0}^{n} {n \choose k} = 2^n.

This problem is a classic example in combinatorics and involves understanding the properties of binomial coefficients. The binomial coefficients, often seen in Pascal's triangle, represent the number of ways to choose a subset of elements from a larger set. The theorem that their sum equals a power of two is not only elegant but also has deep implications in probability, algebra, and computer science.

One strategy to prove this involves using the binomial theorem, which states that for any integer n and any real numbers x and y, (x + y)^n = sum_{k=0}^{n} C(n, k) x^{n-k} y^k. By setting $x = 1$ and $y = 1$, we derive that the sum of binomial coefficients for a given n indeed equals $2^n$. This illustrates the utility of algebraic manipulation in combinatorics.

Another approach to proving this identity is via a combinatorial argument. Consider the set of all subsets of a set with n elements. There are $2^n$ such subsets. The left-hand side of the equation counts the subsets by size, while the right-hand side simply states the total number of subsets. Thus, equating these two sides gives us the desired identity, providing an intuitive understanding of the problem.