Evaluate $\displaystyle\int_V xy^9z^8 (1-x-y-z)^4\,dx\,dy\,dz$ 
Evaluate $\displaystyle\int_V xy^9z^8 (1-x-y-z)^4\,dx\,dy\,dz,$ where $V$ is the pyramidal region $x, y, z \geq 0, x+ y+ z \leq 1.$

So the integral is equivalent to $$\int_0^1\displaystyle\int_0^{1-z} \int_0^{1-y-z} xy^9z^8(1-x-y-z)^4\,dx\,dy\,dz.$$
I can evaluate this directly, which gives the answer $\dfrac{9!8!4!}{25!},$ but that method is way too tedious. Is there a faster approach?
 A: The Beta function is defined by
$$B(p, q)=\int_0^1{x^{p-1}(1-x)^{q-1}}dx \,\, (p, q>0)$$
with a useful property that
$$B(p, q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$
The proof is almost straightforward. Now with this tool we can easily compute integrals of following form:
$$\int_0^a{x^{p-1}(a-x)^{q-1}}dx$$
by substituting $ax$ into $x$ as follows.
$$\int_0^1{(ax)^{p-1}(a-ax)^{q-1}a}dx=a^{p+q-1}\int_0^1{x^{p-1}(1-x)^{q-1}}dx=a^{p+q-1}B(p, q).$$
And your integral is indeed an iteration of calculation of this kind of integral:
$$
\begin{align}
& \int_0^1{\int_0^{1-z}{\int_0^{1-y-z}{(1-x-y-z)^{p-1}x^{q-1}y^{r-1}z^{s-1}}}}dxdydz \\
& =\int_0^1{\int_0^{1-z}{y^{r-1}z^{s-1}\int_0^{1-y-z}{(1-x-y-z)^{p-1}x^{q-1}}}}dxdydz\\
& =\int_0^1{\int_0^{1-z}{(1-y-z)^{p+q-1}y^{r-1}z^{s-1}B(p, q)}}dydz\\
& =B(p, q)\int_0^1{z^{s-1}\int_0^{1-z}{(1-y-z)^{p+q-1}y^{r-1}}}dydz\\
& =B(p, q)\int_0^1{z^{s-1}(1-z)^{p+q+r-1}B(p+q, r)}dydz\\
& =B(p, q)B(p+q, r)B(p+q+r, s)\\
& =\frac{\Gamma(p)\Gamma(q)\Gamma(r)\Gamma(s)}{\Gamma(p+q+r+s)}.\\
\end{align}
$$
