Birgit Anne wrote:Can anybody explain me how 8C3 is calculated? I do not understand this term.
Thank you!
8C3 is a shortform representation of the combinations formula. When we see "8C3" we actually read in our heads "8 choose 3".
The general combinations formula is:
nCk = n!/k!(n-k)!
in which n = the total number of objects available and k = the number of objects that you're selecting.
If you're not familiar with the ! operation, "!" stands for "factorial". We can apply the factorial operation to any non-negative integer and, with the exception of 0!, to calculate we simply multiply the integer noted by all the smaller positive integers. So, for example, :
4! = 4*3*2*1
6! = 6*5*4*3*2*1
and two important factorials to remember:
1! = 1
0! = 1
So, back to 8C3:
8C3 = 8!/3!(8-3)!
= 8!/3!5!
= 8*7*6*5*4*3*2*1/3*2*1*5*4*3*2*1
= 8*7*6/3*2*1
(the 5*4*3*2*1 on the top cancelled out with the 5*4*3*2*1 on the bottom)
= 8*7 = 56
Now, if we know some calculation shortcuts, we can save a bunch of time, which is always good on the GMAT (on a grade 11 math test you might need to show all your work to get 10 out of 10 on a problem, but on the GMAT we just need to click the right circle on the screen!).
So, for factorials, here are a few calculation tricks:
1) never bother writing out the "*1" at the end, since multiplying by 1 does nothing.
2) cancel out the bigger factorial on the bottom with a chunk of the top. Back to our example:
8!/3!5!
Well, we know that 8! = 8*7*6*5!, so we can rewrite our expression as:
8*7*6*5!/3*2*5!
= 8*7*6/3*2
You can always cancel out as shown, saving yourself time. So, if we saw:
10C2
and needed to calculate, we could write is as:
10!/2!8! = 10*9*8!/2!8! = 10*9/2
Hope that helps!
