TITLE: What quantifies as a rigorous proof?
QUESTION [5 upvotes]: Okay I have been thinking about this common combinatorial identity. $$\sum_{r=0}^{n} \binom{n}{r} = 2^n.$$ It is simple to prove this by induction, but it requires some annoying algebraic manipulation which involves other combinatorial identities. What about if we took an approach that takes into factor a specific rephrase of the problem?
My approach is this. Consider a combination lock with n number of on-off switches. To open it require turning on the correct switches. How many combinations are there? clearly, the first switch can be on or off, second on and off etc. so the combintions are $2^n$
We look at it from a different angle. How many ways are there to pick one on switch. $\dbinom{n}{1}$. what about picking 2 switches? $\dbinom{n}{2}$... and so on. Since the number of combinations are supposed to be the same,it gives us the identity.
Now i know this explanation is not rigorous, but can someone point out exactly to me exactly which part of it it fails to be so?
I have also encountered this problem in man other areas of mathematics when I try to prove something, I used a specific problem to do so and thus I doubt the validity of the proof. I can give more examples if needed

REPLY [2 votes]: You can make such an argument rigorous. When things start to get more sophisticated, it often helps immensely to give more detail than one usually sees in this style of proof.
Let $S$ be a set with $n$ elements. Let $P(S)$ be its powerset -- that is, the set of all subsets of $S$. Let $P_i(S)$ be the set of all subsets of $S$ with exactly $i$ elements.
Your proof is then the invocation of four facts:

The well-known formula for the number of elements in $P(S)$
The well-known formula for the number of elements in $P_i(S)$
The observation that $P(S)$ is the disjoint union of all of the $P_i(S)$
The relationship between the number of elements in a sequence of sets and the number of elements in their disjoint union