<s>[INST] Binomial related problem $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = $
Where ${}^n{C_r} = \left( {\begin{array}{*{20}{c}}
n\\
r
\end{array}} \right)$
My approach is as follow $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^5{C_r}.{}^{50 - 10r}{C_5}} $
Using the technique of some previous problem I tried to solve by absorption
$\sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{5!}}{{r!\left( {5 - r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{5!\left( {45 - 10r} \right)!}}}  = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{\left( {45 - 9r} \right)!}}{{r!\left( {45 - 10r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{\left( {5 - r} \right)!\left( {45 - 9r} \right)!}}}  \Rightarrow \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^{45 - 9r}{C_r}.{}^{50 - 10r}{C_{5 - r}}} $
Not able to proceed further
 [/INST] We can generalize the problem in this form:
\begin{align*}
a_{m, n} = \sum_{k \geq 0} (-1)^{n + k} \binom{n}{k}\binom{mk}{n} =\ ?
\end{align*}
For $n = 5$ and $m = 10$, the LHS is the same as the given question. Let
\begin{align*}
f(x) = \sum_{n \geq 0}a_{m, n}x^n
\end{align*}
be the generating function of the sequence $\{a_{m, n}\}_{n \geq 0}$ for a fixed $m$. Then we must find the coefficient of $x^n$ in $f(x)$:
\begin{align*}
f(x) = \sum_{n \geq 0}a_{m, n}x^n &= \sum_{n \geq 0}\left(\sum_{k \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}\right)x^n\\
&= \sum_{n \geq 0}\sum_{k \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}x^n\\
&= \sum_{k \geq 0}\sum_{n \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}x^n\\
&= \sum_{k \geq 0}(-1)^k\binom{n}{k}\sum_{n \geq 0}\binom{mk}{n}(-x)^n\\
&= \sum_{k \geq 0}(-1)^k\binom{n}{k}(1 - x)^{mk}\\
&= \sum_{k \geq 0}\binom{n}{k}\big(-(1 - x)^{m}\big)^k\\
&= \big(1 - (1 - x)^m\big)^n\\
&= \big(1 - (1-x)\big)^n\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n\\
&= x^n\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n
\end{align*}
Therefore, for finding the coefficient of $x^n$ in $f(x)$, we must find the constant coefficient in $\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n$. But the constant coefficient achieves when $x = 0$. So it equals $\big(1 + (1 - 0) + \cdots + (1-0)^{m-1}\big)^n = m^n$. Thus we have
\begin{align*}
\bbox[5px, border: 2px solid blue]{a_{m, n} = \sum_{k \geq 0} (-1)^{n + k} \binom{n}{k}\binom{mk}{n} = m^n}
\end{align*}
So for $n = 5$ and $m = 10$, the result is $10^5$.
</s>