Product of r Choose m with m Choose k/Proof 1

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Theorem

$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$


Proof

Integral Index

Let $r \in \Z$.


Then:

\(\ds \binom r m \binom m k\) \(=\) \(\ds \frac {r^{\underline m} } {m!} \frac {m^{\underline k} } {k!}\)
\(\ds \) \(=\) \(\ds \frac {r! m!} {m! \paren {r - m}! k! \paren {m - k}!}\)
\(\ds \) \(=\) \(\ds \frac {r! \paren {r - k}!} {k! \paren {r - k}! \paren {m - k}! \paren {r - m}!}\)
\(\ds \) \(=\) \(\ds \binom r k \binom {r - k} {m - k}\)

$\Box$


Real Index

Both sides of the above equation are polynomials in $r$.

Since these polynomials agree for all $r \in \Z$, they must agree for all $r \in \R$.

$\blacksquare$


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