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
 A: HINT:
In your expression , what we are looking for  is the sum of the coefficients of the term $x^5$ in the expansion of $$-\bigg[\bigg(1-(1+x)^{10}\bigg)^5 -1\bigg]$$
Because :

*

*$$[x^5][\bigg(1-(1+x)^{10}\bigg)^5 =\sum_{k=0}^{5}-\binom{5}{0}[(1+x)^{10}]^5+\binom{5}{1}[(1+x)^{10}]^4-\binom{5}{2}[(1+x)^{10}]^3...+\binom{5}{5}[(1+x)^{10}]^0$$
To render this expression into yours , subtract the last term , and multiply by $(-1)$
A: Consider the $50$ (ordered) pairs of numbers $(a,b)$, where $a\in\{1,2,3,4,5\}$ and $b\in \{0,1,2,3,\ldots, 9\}$.
Let $S$ be all possible $5$-combinations of the $50$ pairs, without restrictions. Then
$$|S| = \binom{50}{5}$$
Let $A_i$ be the subset of combinations in $S$ which has no pairs $(i,b)$, i.e. $5$-combinations with none of $(i,0)$, $(i,1)$, $(i,2)$, ..., and $(i,9)$. The number of such combinations is
$$|A_i| = \binom{50-10}{5} = \binom{40}5$$
Similarly, $A_i\cap A_j$ would the the subset of combinations in $S$ which has none of pairs of form either $(i,b)$ or $(j,b)$. For $i\ne j$, the number of such combinations is
$$|A_i\cap A_j| = \binom{50-20}{5} = \binom{30}5$$
With reference to the complementary form of the inclusion-exclusion principle, consider the intersection $\bigcap_{i=1}^5 \overline{A_i}$,
$$\begin{align*}
\left|\bigcap_{i=1}^5 \overline {A_i}\right|
&= |S| - \sum_{i=1}^5 |A_i| + \sum_{1\le i<j\le 5} |A_i\cap A_j| - \cdots + (-1)^5 |A_1\cap A_2\cap A_3\cap A_4\cap A_5|\\
&= \binom{50}{5} - \sum_{i=1}^5 \binom{40}{5} + \sum_{1\le i<j\le 5} \binom{30}{5} -\cdots + (-1)^5\binom{0}{5}\\
&= \binom{5}{0}\binom{50}{5} - \binom{5}{1}\binom{40}{5} + \binom{5}{2}\binom{30}{5} - \binom{5}{3}\binom{20}{5} + \binom{5}{4}\binom{10}{5} - \binom{5}{5}\binom{0}{5}\\
&= \binom{5}{0}\binom{50}{5} - \binom{5}{1}\binom{40}{5} + \binom{5}{2}\binom{30}{5} - \binom{5}{3}\binom{20}{5} + \binom{5}{4}\binom{10}{5}
\end{align*}$$
Alternatively, the intersection $\bigcap_{i=1}^5 \overline{A_i}$ is the subset of combinations which has some pairs $(1,b_1)$, some pairs $(2,b_2)$, ... , and some pairs $(5,b_5)$, forming a combination of $5$ pairs. Which means there is exactly one pair $(a,b_a)$ for each $a\in\{1,2,3,4,5\}$.
Decimal strings of $5$ digits (or decimal integers possibly with leading $0$) correspond 1-to-1 to this subset, so
$$\left|\bigcap_{i=1}^5 \overline {A_i}\right| = 10^5 = 100\,000$$
A: 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 $10^5$.
