How do I do

13!
---------
(10-4)!4!

The part that confuses me is the (6!)4!

When in doubt, write out the terms for each factorial and cancel.


13*12*11*10*9*8*7*6*5*4*3*2*1
---------------------------------------------
(6*5*4*3*2*1)(4*3*2*1)

This might take a little extra time, but you can see exactly what cancels and what's left over. Hope this helps.

what do you mean "do"? just leave it be.

Supposedly the equation is equal to 286, but I don't know how to come up with that number.

are you thinking of combinations, 13 choose 10? or similarly 13 choose 3? that's equal to 286, but that's not what you wrote. yours is not equal to 286

13 choose 10 = 13!/((13-10)!10!)
13 choose 3 = 13!/((13-3)!3!)

which are both equal since:
n choose k = n choose (n-k)

This post was edited by carteblanche on Apr 20 2014 09:44pm

Yes! How would I do that?

same way dude above showed you.

13!
_______
10!3!



13*12*11*10!
_____________
10!3!

since 10 is very close to 13, simplification is very helpful. just cross out the 10! on each side

13 * 12 * 11
____________
3 * 2 * 1

more simplification since 3 * 2 are both factors of 12:

13 *2 * 11 = 286

This post was edited by carteblanche on Apr 20 2014 09:48pm

I appreciate the help guys, but it still isn't clicking in my mind

which step are you not getting? keep in mind that factorial is just multiplication. are you familiar with the summation formula?



factorials are similar, the difference is that instead of adding you're multiplying. you can represent it with pi instead of sigma iirc



in much the same way that [sum of i, from i = 1 to n] = 1 + 2 + 3 + ... + n

n! is [product of i, from i = 1 to n] = 1 * 2 * 3 * ... * n

This post was edited by carteblanche on Apr 20 2014 10:11pm

if youre forced to simplify these by hand youll likely have to write it out first write it out fully:

13*12*11*10*9*8*7*6*5*4*3*2*1
---------------------------------------------
(4*3*2*1) (6*5*4*3*2*1)


you can see both the numerator and denominator have a few same numbers in common (the 1*2*3*3*4*5*6) and they both have the same operator so they will cancel to get:

13*12*11*10*9*8*7
---------------------------------------------
(4*3*2*1)

now simplify again 2*4=8 so you can cancel a 8 from both numerator and denominator to get:

13*12*11*10*9*7
---------------------------------------------
(3*1)

1*3 is just 3 so the 1 goes away to get :

13*12*11*10*9*7
---------------------------------------------
3

now you can divide the 9 by the 3 to get:

13*12*11*10*3*7

which is simple multiplication