how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?

Question

I have done the following transformation:

$(1-x)^{m-1}=u$ $\Rightarrow -(m-1)(1-x)^{m-2}dx=du$

$x^{n-1}dx=dv$ $\Rightarrow \frac{x^n}{n}=v$

Hence $$\int^1_0x^{n-1}(1-x)^{m-1}dx= (1-x)^{m-1}\frac{x^n}{n}\bigg |^1_0+ \int\frac{x^n}{n}(m-1)(1-x)^{m-2}dx=(1-1)^{m-1}\frac{1}{n}-1^{m-1}\frac{0}{n} + \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx= \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx=\frac{m-1}{n}B(n+1,m-1)$$

But i'm not exactly sure how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?

Answer 1

Beginning with the result you have shown, repeated application yields the following chain of equalities:

\begin{align*} B(n,m) &= \frac{m-1}{n}\cdot B(n+1,m-1) \\ &= \frac{m-1}{n}\cdot \frac{m-2}{n+1}\cdot B(n+2,m-2) \\ &= \cdots \\ &= \frac{m-1}{n}\cdot \frac{m-2}{n+1} \cdot \cdots \cdot \frac{1}{n+m-2} \cdot B(n+m-1,1) \\ &= \frac{(m-1)!(n-1)!}{(n+m-2)!}\cdot B(n+m-1,1) \end{align*}

It remains to calculate $B(n+m-1,1)$: $$B(n+m-1,1) = \int_0^1x^{n+m-2} \, \mathrm{d}x = \left[ \frac{x^{n+m-1}}{n+m-1} \right]_0^1 = \frac1{n+m-1}.$$

Hence, the result follows, namely: $$B(n,m) = \frac{(m-1)!(n-1)!}{(n+m-1)!}$$

Answer 2

Start by showing \begin{eqnarray*} B(n+m-1,1)= \frac{1}{n+m-1}. \end{eqnarray*} Now use induction to prove \begin{eqnarray*} B(n+m-j,j)= \frac{(j-1)!}{(n+m-1)(n+m-1)\cdots(n+m-j)}. \end{eqnarray*}