# Show that p choose n=0 when n>p

1. May 24, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Show that if p is a positive integer, then $\binom{p}{n}=0$ when $n>p$, so $(1+x)^p=\sum \binom{p}{n} x^n$, is just a sum of $p+1$ terms, from $n=0$ to $n=p$. For example $(1+x)^2$ has 3 terms, $(1+x)^3$ has 4 terms, etc. This is just the familiar binomial theorem.

2. Relevant equations
3. The attempt at a solution

$$\binom{p}{n}=\frac{p(p-1)(p-2)\cdots(p-n+2)(p-n+1)}{n!}$$ Since n>p I figured I would test with (p+1) to see what happens. $$\binom{p}{p+1}=\frac{p(p-1)(p-2)\cdots(p-(p+1)+2)(p-(p+1)+1)}{(p+1)!}$$ So the last term clearly goes to zero. Is rewriting $\binom{p}{n}=0$ as $$\frac{(p-n+1)(p-n+2)(p-n+3)(p-n+4)\cdots p}{n!}$$ sufficient for proving this? I am also quite lost as to what they're implying with $$(1+x)^p=\sum \binom{p}{n} x^n$$ I understand that there is always $(p+1)$ terms after expanding a binomial raised to a power but by the sum are they referring to if the binomial was rewritten as a Maclaurin series?

2. May 25, 2015

### Svein

You are so close! You have already $\begin{pmatrix} p\\ n\\ \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-n+1)}{n!}$. Therefore $\begin{pmatrix} p\\ p\\ \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-p+1)}{p!}=1$ and $\begin{pmatrix} p\\ p+1\\ \end{pmatrix}=\frac{p(p-1)(p-2)\dotso (p-(p+1)+1)}{(p+1)!}=0$. So, for n≥(p+1), you have one factor = 0, therefore the product is 0.