This problem is equivalent to determining $a$, $b$, $r$, $s$, and $t$ such that the given relation is an identity of integer $n$. We claim $a=b=2$ is the only solution. Accordingly, $r=3$, $t=4$ and $s=1$.

First, let's shown $a=1$ or $b=1$ will lead to no solution.

When $a=1$, the equation can be simplified into $$ \frac{n(n-1)\cdots(n-b+1)}{b!}=\frac{r(n+s)(n+s-1)\cdots(n+s-t+1)}{t!}$$

There are $b$ terms on the left and $t$ terms on the right, hence, we must have $b=t$, it follows $r$ must be $1$ by comparing the coefficient of the highest powered term. It follows that $$n(n-1)\cdots(n-b+1)=(n+s)(n+s-1)\cdots(n+s-b+1)$$

This relation can only hold when $s=0$. But $s$ is a positive integer. Hence, it is a contradiction.

Similarly, when $b=1$, the left side becomes $C_n^a$. The same reasoning will lead to contradiction as well.

Now, let's show $a > 2$ or $b > 2$ will lead to no solution as well. 

For this, let $$N=\binom{n}{a}=\frac{\cdot n(n-1)\cdots(n-(a-1))}{a!}$$

The left side of the equation becomes $$\frac{1}{b!}\cdot N(N-1)(N-2)\cdots(N-(b-1))$$

Meanwhile, the right side is a product of a constant coefficient and a series of consecutive terms.

$$\frac{r}{t!}\cdot (n+s)(n+s-1)\cdots(n+s-t+1)$$

In order to make the left always equal the right, every term on the left must be factorizable into a product of several one-degree polynomials whose constant terms are integers. Additionally, these constant terms must be distinct because there is no duplicate terms on the right.

The first term $N$ is already a product of consecutive terms by definition. Let's consider the second term $(N-1)$. This is equivalent to determining integers $w_1$, $w_2$, $\cdots$, $w_a$ such that the following relation holds after taking out the constant coefficient: $$n(n-1)\cdots(n-(a-1)) - a! = (n-w_1)(n-w_2)\cdots(n-w_a)$$

By rational root theorem, all these $w_i$ must divide $a!$. Meanwhile, they must be distinct and not in $\{0, 1, \cdots, a-1\}$ because taking any of these values will cause duplication with $N$'s divisors. This will force these $w_i$ to take $-1$, $-2$, $\cdots$, $-(a-1)$ and one of $\pm a$ depends on the odd-even parity of $a$.

Furthermore, let's show $a$ cannot be greater than $2$, i.e. it must be in the form of $$n(n-1)-2 = (n-w_1)(n-w_2)=(n-2)(n+1)$$

If there are more than two terms, let's consider the coefficient of the term of the second highest power, i.e. $n^{a-1}$. On the left, it is $$(-1) + (-2) + \cdots -(a-1)$$

On the right, it will be $$1+2 +\cdots (a-1)\pm a$$

This cannot hold. Therefore $a=2$ is necessary for this problem to be solvable. At this time the left side will be $$N(N-1)\cdots = n(n-1)\cdot (n-2)(n+1) \cdots$$

Now, if $b > 2$, then the term $(N-2)$ will exist. We will then need to determine integers $u_1$, $u_2$, $u_3$ such that $$n(n-1) - 2\times 2! = (n-u_1)(n-u_2)$$

This is impossible. Therefore $b$ has to equal $2$ in order to make this problem solvable.

When $a=b=2$, there indeed exists a solution. Therefore, we conclude that $(a, b)=\boxed{(2, 2)}$ is the only solution.