Q:

How to calculate the sum of combinatorial numbers

For my work on an almost completely unrelated field I came across the following formula. I know that I have learned that all in high school. But since this is more than 15 years ago in which I never used this kind of mathematics I now realize that I am completely lost...
So I have the following formula:
$$\sum_{A=42}^{200} \binom{A-1}{A-42} \times 6^{42}$$
I know the first step to calculate this is the following:
$$\sum_{A=42}^{200} \frac{(A-1)!}{(A-42)! \times ((A-1)-(A-42))!} \times 6^{42}$$
Now I need the solution to this formula in terms of $10^x$, so basically the $x$ value. What I am interested in is of course also the steps of this calculation, to understand how this is done.

A:

You can use something known as the Hockeystick Identity:
$\displaystyle\sum_{A = 42}^{200}\dbinom{A-1}{A-42} = \sum_{A = 42}^{200}\dbinom{A-1}{41} = \sum_{n = 41}^{199}\dbinom{n}{41} = \dbinom{199+1}{41+1} = \dbinom{200}{42}$. 
Now, multiply both sides by $6^{42}$. 
Using WolframAlpha, we get that $\dbinom{200}{42}6^{42} \approx 1.46 \times 10^{76}$. If you wanted to evaluate this by hand, you can write the sum as $\dfrac{200! \cdot 6^{42}}{158! \cdot 42!}$ and then use Sterling's Approximation.

