New Year Maths 2015

In the spirit of the festive period and in appreciation of the encouraging response to my Xmas Combinatorics 2014 problem posted recently, here's one for the New Year!

Express the following as a product of four binomial coefficients: $$\color{red}{\sum_{r=1}^{M-1}r}\color{orange}{\sqrt{\int_0^X 2x dx}} \color{green}{\left(\prod_{i=1}^{120}(V!-i+1) \right)} \color{indigo}{\left(\prod_{h=1}^Mh\right)} \left[\color{maroon}{\left(\prod_{n=1}^{120}n\right)}\color{blue}{\left(\prod_{y=1}^M y\right)}\right]^{-1} $$

$$\begin{align} &\color{red}{\sum_{r=1}^{M-1}r}{\color{orange}{\sqrt{\int_0^X 2x dx}}} \color{green}{\left(\prod_{i=1}^{120}(V!-i+1) \right)} \color{indigo}{\left(\prod_{h=1}^Mh\right)} \left[\color{maroon}{\left(\prod_{n=1}^{120}n\right)} \color{blue}{\left(\prod_{y=1}^M y\right)}\right]^{-1} \\\\ &=\color{orange}{\sqrt{\int_0^X 2x dx}} \color{red}{\sum_{r=1}^{M-1}r} \color{green}{\left(\prod_{i=1}^{120}(V!-i+1) \right)} \color{indigo}{\left(\prod_{h=1}^Mh\right)} \left[\color{maroon}{\left(\prod_{n=1}^{120}n\right)} \color{blue}{\left(\prod_{y=1}^M y\right)}\right]^{-1} \\\\ &=\color{orange}{\sqrt{\left[{x^2}\right]_0^X}}\cdot \color{red}{\frac{M(M-1)}2}\cdot \color{green}{V!^{\underline{120}}} \color{indigo}{M!} \left[\color{maroon}{120!}\color{blue}{M!}\right]^{-1} \\\\ &=\color{orange}X\cdot \color{red}{\frac{M(M-1)}2}\cdot \color{green}{V!^{\underline{120}}} \color{indigo}{M!} \left[\color{maroon}{120!}\color{blue}{M!}\right]^{-1} \\\\ &=\color{orange}{\frac X1}\cdot \color{red}{\frac{M(M-1)}{1\cdot 2}}\cdot \color{green}{\frac{V!^{\underline{120}}}{\color{maroon}{120!}}}\cdot \color{indigo}{\frac{M!}{\color{blue}{M!}}}\\\\ &=\color{orange}{\binom X1} \color{red}{\binom M2} \color{green}{\binom {V!}{120}} \color{indigo}{\binom MM}\\\\ &=\color{red}{\binom M2} \color{indigo}{\binom M0} \color{orange}{\binom X1} \color{green}{\binom {V!}{5!}}\\\\ \\\\ \end{align}$$

Happy New Year!