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Can someone using only these conditions

  1. $a_{m,k}=a_{m-1,k}+a_{m-1,k-1},m>k$

  2. $a_{m,k}=1,m=k$

  3. $a_{m,k}=0,m

prove that

$a_{m,k}=\frac{m!}{k!(m-k)!}$

here is way to construct Pascal triangle. I am interested to view direct proof. I read many books on combinatorics and nowhere such proof is done yet.

  • 1
    Let $b_{m,k}=\frac{m!}{k!(m-k)!}$. We need to verify that $b_{m,k}$ satisfies same recurrence, initial conditions as $a_{m,k}$. This is a calculation.2011-12-04

1 Answers 1

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Hints: consider the formal series $a_m(x)=\sum\limits_{k}a_{m,k}x^k$. Show that $a_m(x)=\sum\limits_{k=0}^ma_{m,k}x^k$ for every $m\geqslant0$. Show that $a_0(x)=1$ and $a_m(x)=(1+x)a_{m-1}(x)$ for every $m\geqslant1$. Deduce that, for every $m\geqslant0$, the relation $a_m(x)=(1+x)^m$ holds, and finally, the value of every $a_{m,k}$.