Pascal's triangle, estimate row value by fixed row and maximum yields. let res be the max yields. L

Pascal's triangle, estimate row value by fixed row and maximum yields.
let res be the max yields. Let col be a fixed positive integer. Let $f(row)=(\genfrac{}{}{0ex}{}{\text{row}}{\text{col}})=A$, what is the largest value for row such that $f(row)\le <!--\; \le \; -->A$?
Standard Pascal's triangle expression:
$(\genfrac{}{}{0ex}{}{\text{row}}{\text{col}})=\frac{\text{row}!}{\text{col}!(\text{row-col})!}=\text{result}$
Update
I am archiving a regression program, which takes raw data of two-dimensional vertexes, and return a function correspond to the rule. It can be called which pass in x and return a y value.
In this procedure, generator yields different amount of sample data from the raw data. For example, in simplest linear regression
y=kx+b
which takes a minimum points of 2, [x1, y1], [x2, y2], for working out the coefficients k and b.
Here comes the part which is the most relevant to this question.
if the raw data contains n vertexes, then the amount of iteration equal to
$(\genfrac{}{}{0ex}{}{\text{n}}{\text{required}})=\frac{\text{n}!}{\text{required}!(\text{n-required})!}=\text{yields}$
for instance, for linear regression, required is equal to 2. for 10 dots:
$(\genfrac{}{}{0ex}{}{\text{10}}{\text{2}})=\frac{\text{10}!}{\text{2}!(\text{10-2})!}=\text{45}$
However, the original design of my program is to iterate all through the amount of yields, for example, when n=10, required=2 which res=45, do an iteration of 45.
The problem is:
n<20> required<2>: res<190>
n<40> required<2>: res<780>
n<60> required<2>: res<1770>
n<80> required<2>: res<3160>
n<100> required<2>: res<4950>
n<120> required<2>: res<7140>
n<140> required<2>: res<9730>
n<160> required<2>: res<12720>
n<180> required<2>: res<16110>
when n>=16, it freezes and overloads the machine, the fan is very loud.
Therefore, I seek for a way,
with known required, by limiting the largest amount of iteration, how many dots do I need to extract from raw data?
which the question can be briefed to:
$(\genfrac{}{}{0ex}{}{\text{n}}{\text{required}})=\frac{\text{n}!}{\text{required}!(\text{n-required})!}=\text{yields}$
with known the regression takes required vertexes, set a maximum iteration of yields, find a suitable n value.
Thanks very much for reading till the end. I am sorry if anything I have written doesn't make sense to you. Any help is appreciable.