# Pascal’s rule proof

We need to show

 $\displaystyle\binom{n}{k}+\binom{n}{k-1}$ $\displaystyle=$ $\displaystyle\binom{n+1}{k}$

Let us begin by writing the left-hand side as

 $\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-(k-1))!}$

Getting a common denominator and simplifying, we have

 $\displaystyle\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}$ $\displaystyle=$ $\displaystyle\frac{(n-k+1)n!}{(n-k+1)k!(n-k)!}+\frac{kn!}{k(k-1)!(n-k+1)!}$ $\displaystyle=$ $\displaystyle\frac{(n-k+1)n!+kn!}{k!(n-k+1)!}$ $\displaystyle=$ $\displaystyle\frac{(n+1)n!}{k!((n+1)-k)!}$ $\displaystyle=$ $\displaystyle\frac{(n+1)!}{k!((n+1)-k)!}$ $\displaystyle=$ $\displaystyle\binom{n+1}{k}$
Title Pascal’s rule proof PascalsRuleProof 2013-03-22 11:47:14 2013-03-22 11:47:14 akrowne (2) akrowne (2) 10 akrowne (2) Proof msc 05A10 msc 81T13 msc 53C80 msc 82-00 msc 83-00 msc 81-00