why is ${n+1\choose k} = {n\choose k} + {n\choose k-1}$? [duplicate]

Question

Can someone explain to me the proof of $${n+1\choose k} = {n\choose k} + {n\choose k-1}$$?

Answer 1

I'll just be completely unimaginative here:

$$\eqalign{{n\choose k}+{n\choose k-1}&= {n!\over (n-k)!k!}+ {n!\over (n-(k-1))! (k-1)!}\cr &={n!\over (n-k)!k!}+ {n!\over (n-k+1)! (k-1)!}\cr &={(n-k+1)n!\over(n-k+1) (n-k)!k!}+ {n!k\over (n-k+1)! (k-1)! k}\cr &={(n-k+1)n! + n!k\over (n-k+1)! k!}\cr &={n\cdot n !-kn!+n!+n!k\over (n-k+1)! k!}\cr &={n\cdot n ! +n! \over (n-k+1)! k!}\cr &={n!(n+1) \over (n-k+1)! k!}\cr &={(n+1)! \over( (n+1)-k)! k!}\cr &={n+1\choose k}. }$$

Answer 2

One proof goes like this: Suppose you have a list of all $\dbinom {n}{k-1}$ ways to choose $k-1$ objects out of $n$. Then we add an $(n+1)^\text{th}$ object to those from which we can choose. How do we make a list of all $\dbinom{n+1}{k}$ ways to choose $k$ out of these $n+1$? Here's how: first make a list of all ways to choose $k$ out of the original $n$ objects. That's a partial list. All of its items exclude the "new" object. It has $\dbinom nk$ items. Then take the list you already had, of all ways to pick $k-1$ out of those $n$. To each item on the list, giving $k-1$ objects, you add the "new" object, getting a set of $k$ of the $n+1$. That's another partial list. All of its items include the "new" object. It has $\dbinom{n}{k-1}$ items.

Now just add the two partial lists together: the list of those that exclude the "new" object—there are $\dbinom nk$ of those—and the list of those that include the "new" object—there are $\dbinom{n}{k-1}$ of those.

Answer 3

The identity is true by vacuity, because that's how my teacher defines a binomial coefficient. [Actually, one needs to append the base conditions to that for a complete definition.]

On a serious note, please define the notation that you use. As I demonstrated just now, an otherwise interesting question could become trivial for such mundane reasons as differing definitions or conventions.