You can cross out equal factors in numerator and denominator. For example:
$$\frac{6!}{4! \times2!} = \frac{6\times5\times4\times3\times2\times1}{(4\times3\times2\times1) (2\times1)} $$
Now cross out all similar factors in bottom and top. Only one factor at a time in top and bottom. Leaving the following:
$$ \frac{6\times5\times4\times3\times2\times1}{(4\times3\times2\times1) (2\times1)} = \frac{6\times5} {2\times1} = \frac{3\times5} {1} = 3\times5 = 15 $$
So for your problem it will still be a hassle. I'm sure you can factor out even more. Did you try a calculator, or are you getting an overflow?
You can perhaps solve it with programming. This is an R script for the calculation:
total = 1
for (i in 949:899){
total = total * i
}
total/factorial(50)
It outputs 5.79942e+86