# Can the Binomial Theorem be proved probabilistically?

I believe that the answer is affirmative, as there is a reputable-appearing subscription-hidden link on the web claiming such, and because, although I have not seen that proof, I have (I believe) a proof of my own (which I’m pretty sure is similar to the hidden one).

Here is the link to the subscription-hidden article.

And here’s my take on the question that I have posed:

Using elementary Probability Theory, and an appeal to symmetry and continuity, it is easy to prove the Binomial Theorem. First, we note that it is easy to establish that $\binom{n}{k}$ = C(n,k), by first (easily) establishing the formula for P(n,k), and then (easily) modifying it to give C(n,k). We can easily show that the binomial coefficients must be C(n,k) for the expansion of ${(x + y)}^n$, for positive integers x and y, by noting the following situation: If an urn contains x white balls and y black balls, then, given that (exactly) $n$ balls are selected, with replacement, the events: 0. picking n white balls 1. picking n – 1 white balls and 1 black ball 2. picking n – 2 white balls and 2 black balls … n – 1. picking 1 white ball and n – 1 black balls n. picking n black balls are mutually-exclusive and exhaustive events, and so the sum of their probabilities is unity. However, it is obvious that the number of ways in which event k can occur is C(n,k). Since all the terms have a denominator of ${(x + y)}^n$, we can multiply both sides of the equation by ${(x + y)}^n$ to get the Binomial Theorem. The case for arbitrary integers x and y follows by symmetry, and the case for arbitrary real numbers x and y follows by continuity. Done.

Notation: $\binom{n}{k}$ is $\frac{n!}{(k!)(n – k)!}$, C(n,k) is the number of combinations of n things taken k at a time, and P(n,k) is the number of permutations of n things taken k at a time.

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I hate to nitpick, but you are using the word "easy" and "obvious" far too many times. I also don't understand exactly what statement you're referring to as the binomial theorem, nor do I understand what distinction you appear to be making between this proof and the usual proof. –  Qiaochu Yuan May 8 '11 at 1:42