Generalization of Binomial Coefficients: Beta Function
Theorem: Binomial Coefficients Expressed Through the Beta Function
$0 \le k\le n$ is satisfied for two natural numbers $k,n$, the following equation holds. $$ \binom{n}{k}={}_{n}C_{k}=C(n,k)=\frac{1}{(n+1)B(n-k+1,k+1)} $$ For two natural numbers $m,n$, the following equation holds. $$ B(m,n)=\left[ \frac{mn}{m+n} \begin{pmatrix} m+n \\ n \end{pmatrix}\right]^{-1} $$
Description
The beta function, defined by $B(p,q):=\displaystyle \int_{0}^{1}t^{p-1}(1-t)^{q-1}dt$, can also be seen as a generalization of binomial coefficients. The proof is not difficult, but a lemma has to be proved first.
Proof
Lemma 1
$$ B(p,q)=B(p+1,q) +B(p,q+1) $$
Proof of Lemma 1
$$ \begin{align*} B(p+1,q) + B(p,q+1) =& \int_{0}^{1} t^{p-1} (1-t)^{q-1}dt + \int_{0}^{1}t^{p-1}(1-t)^{p-1}dt \\ =& \int_{0}^{1}t^{p-1}(1-t)^{q-1}\big[t+(1-t) \big]dt \\ =& \int_{0}^{1}t^{p-1}(1-t)^{q-1}dt \\ =& B(p,q) \end{align*} $$
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Lemma 2
- (a): $B(p+1,q)=\dfrac{p}{p+q}B(p,q)$
- (b): $B(p,q+1)=\dfrac{q}{p+q}B(p,q)$
Proof of Lemma 2
$$ \begin{align*} B(p+1,q) =& \int_{0}^{1}t^{p}(1-t)^{q-1}dt \\ =& \left[ -\frac{1}{q}t^{p}(1-t)^{q}\right]_{0}^{1} + \int_{0}^{1} \frac{p}{q}t^{p-1}(1-t)^{q}dt \\ =& 0 + \frac{p}{q}\int_{0}^{1} t^{p-1}(1-t)^{q}dt \\ =& \frac{p}{q} B(p,q+1) \end{align*} $$ Integration by parts was used for the second equality. Since by Lemma 1, $(a)$, substituting this into the equation yields: $$ B(p+1,q)=\frac{p}{q}B(p,q) -\frac{p}{q}B(p+1,q) \\ \Rightarrow \frac{q+p}{q}B(p+1,q)=\frac{p}{q}B(p,q) \\ \Rightarrow B(p+1,q)=\frac{p}{p+q}B(p,q) $$ Since by Lemma 1, $(b)$, substituting this into the equation yields: $$ B(p,q)-B(p,q+1)=\frac{p}{q}B(p,q+1) \\ \Rightarrow B(p,q)=\frac{p+q}{q}B(p,q+1) \\ \Rightarrow B(p,q+1)=\frac{q}{p+q}B(p,q) $$
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Main Proof
First, let’s show that $B(1,1)=1$. This can be directly seen from the definition. $$ B(1,1)=\int_{0}^{1}t^{0}(1-t)^{0}dt=1-0=1 $$ Let’s say $m,n \in \mathbb{N}$. Applying Lemma 2’s $(a)$ repeatedly to $B(m,n)$ yields: $$ \begin{align*} B(m,n) =& \frac{m-1}{m+n-1}B(m-1,n) \\ =& \frac{m-1}{m+n-1}\cdot \frac{m-2}{m+n-2}B(m-2,n) \\ =& \frac{m-1}{m+n-1} \cdot \frac{ m-2 }{m+n-2}\cdot \cdots \frac{ 1 }{m+n-(m-1) }B(1,n) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) }B(1,n) \end{align*} $$ Applying Lemma 2’s $(b)$ repeatedly results in: $$ \begin{align*} B(m,n) =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) }B(1,n) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }B(1,n-1) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }\cdot \frac{ n-2 }{ n-1 }B(1,n-2) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }\cdot \frac{ n-2 }{ n-1 }\cdots \frac{ 1 }{ n+1-(n-1) }B(1,1) \\ =& \frac{ (m-1)!(n-1)! }{ (m+n-1)(m+n-2)\cdots(n+1)n(n-1)\cdots2 } B(1,1) \\ =& \frac{ (m-1)!(n-1)! }{ (m+n-1)! } \\ =& \frac{ m+n }{ mn }\frac{ m!n! }{ (m+n)! } \\ =& \left[ \frac{ mn }{ m+n } \begin{pmatrix} m+n \\ n \end{pmatrix} \right]^{-1} \end{align*} $$ Substituting $m=n-k+1$, $n=k+1$ into the third equation from the bottom of the above sequence gives: $$ B(n-k+1,k+1)=\frac{(n-k)!k!}{ (n+1)! }=\frac{ (n-k)!k! }{(n+1)n! } $$ Therefore: $$ \frac{ n! }{(n-k)!k! }=\begin{pmatrix} n \\ k \end{pmatrix}=\frac{ 1 }{ (n+1)B(n-k+1,k+1) } $$
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