Question 548996
We know that n!=n*(n-1)*(n-2)*...*2*1

So we can expand 
{{{(11! 12!)/(8!10!)}}}
to get:
{{{(11*10*9*8*7*6*5*4*3*2*1*12*11*10*9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1*10*9*8*7*6*5*4*3*2*1)}}}
Now... looking at the expression above... can anything divide out (i.e. get rid of any number that appears in the numerator and denominator).  This leaves:
{{{(11*12*11*10*9)}}}
Now this expression can be calculated as an integer on your calculator... I will leave that for you to do . 

Obviously if we had large numbers... it wouldn't be a good idea to write out the whole factorial, but you may be able to see some patterns once you do some examples... (try to find them)

Hopefully this helps a bit... let me know if you are still unsure