If 

{{}^{36}C}^{r+1}=\frac{6\cdot {}^{35}C_r}{k^2-\mathrm{3 }}

\frac{\mathrm{}}{} such that S={(r,k)} then number of elements in set S is

The correct option is 4.

We are given the equation:
${}^{36}C_{r+1}=\frac{6\cdot {}^{35}C_r}{k^2-3}$
where we need to find the number of elements in the set S = {(r, k)} containing integer solutions for r and k.

First, let us recall the standard identity for combinations:
${}^{n}C_y=\frac{n}{y}\cdot {}^{n-1}C_{y-1}$

Applying this formula to the left-hand side of the given equation with n = 36 and y = r + 1, we get:
${}^{36}C_{r+1}=\frac{36}{r+1}\cdot {}^{35}C_r$

Now, substitute this expression back into the original equation:
$\frac{36}{r+1}\cdot {}^{35}C_r=\frac{6}{k^2-3}\cdot {}^{35}C_r$

For the combination terms to be defined and non-zero, we must have:
$0\le r\le 35$
where r is a non-negative integer.

Since ${}^{35}C_r\ne 0$, we can divide both sides by it:
$\frac{36}{r+1}=\frac{6}{k^2-3}$

Simplifying the equation by dividing both sides by 6:
$\frac{6}{r+1}=\frac{1}{k^2-3}$
Taking the reciprocal of both sides:
$k^2-3=\frac{r+1}{6}$
Adding 3 to both sides:
$k^2=3+\frac{r+1}{6}$

For k to be an integer, $k^2$ must be a perfect square. Also, since r is an integer, $\frac{r+1}{6}$ must be an integer. This implies that (r + 1) must be a multiple of 6.

Given the range $0\le r\le 35$, the possible values for (r + 1) are:
$r+1\in \{6,12,18,24,30,36\}$

Let us evaluate $k^2$ for each case:
1. If $r+1=6\Rightarrow k^2=3+1=4\Rightarrow k=\pm 2$ (Integers). Here, r = 5. This gives two solutions: (5, 2) and (5, -2).
2. If $r+1=12\Rightarrow k^2=3+2=5$ (No integer solution for k).
3. If $r+1=18\Rightarrow k^2=3+3=6$ (No integer solution for k).
4. If $r+1=24\Rightarrow k^2=3+4=7$ (No integer solution for k).
5. If $r+1=30\Rightarrow k^2=3+5=8$ (No integer solution for k).
6. If (Integers). Here, r = 35. This gives two solutions: (35, 3) and (35, -3).

Thus, the set S containing all such pairs (r, k) is:
$S=\left\{\right(5,2),(5,-2),(35,3),(35,-3\left)\right\}$

The total number of elements in set S is 4.