Why, conceptually, is it that $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$? Why, conceptually, is it that $$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}?$$ I know how to prove that this is true, but I don't understand conceptually why it makes sense.
 A: The number of the combination that we choose $r$ people from $n$ people is the sum of the number of the combination that a person named $A$ is included and the number of the combination that $A$ is not included.
A: Sure.  You're dividing into 2 cases.  On the one hand you're saying I'm stuck choosing this element over here.  So now I have r-1 more choices to make out of n-1 things.  In the other case you're refusing that element.  Now you've eliminated a choice, but still must pick r elements.  These two cases are exhaustive and exclusive.
A: Another simple way to look at it is Pascal's triangle.  This is simply a formulation of the rule that governs the triangle -- an entry is the sum of the two entries above it.  The $i$-th entry in the $n$-th row is the sum of the $i-1$-th entry in the $n-1$-th row and the $i$-th entry in the $n-1$-th row.  Hence,
$$\binom{n-1}{i-1}+\binom{n-1}{i}=\binom{n}{i}$$
A: Consider picking $r$ objects out of a set of $n$ of them. Pick a particular object $O$. Either your set has $O$, or it doesn't. There are $\binom{n - 1}{r - 1}$ sets with $O$, and $\binom{n - 1}{r}$ sets without it. (Do you see why those are true?) So together, there are $\binom{n - 1}{r - 1} + \binom{n - 1}{r}$ sets of size $r$.
A: Because the number of ways that we can make a team of a certain size is equal to the number of ways that we can make a team of the same size without Fred, plus the number of ways we can make a team of size one smaller, plus Fred.
A: The number of ways of choosing $r$ items from $n$ distinct items is the same as 


*

*choosing $r-1$ items from the first $n-1$ items (and add to the last item), plus

*choosing $r$ items from the first $n-1$ items (and not choosing the last item)

