anon1 wrote:

I'm sorry what is the C standing for when you guys write something like 30C28 or 29C27 ?

I'll also suggest this blog which discusses the idea.

http://magoosh.com/gmat/2012/gmat-permu ... binations/The basic idea is that "30C28" (read as "thirty choose twenty-eight") is the number of combination of 28 items we could select from a pool of 30 unique items. For simplicity, let's pick a simpler number.

5C3 would be the number of combinations of three I could pick from a pool of five unique items. As it happens, 5C3 = 10. Think about the set {A, B, C, D, E}, which has five unique items. Here is the set of 10 possible combinations of three

A, B, C

A, B, D

A, B, E

A, C, D

A, C, E

A, D, E

B, C, D

B, C, E

B, D, E

C, D, E

Notice, we are counting "combinations" in which order doesn't matter, so ABC and CAB would count as the same combination. When order does matter, that's something called "permutation", which is also discussed in that blog article.

To evaluate a number like 6C2 -----

a) some calculators have nCr as a function, but of course, you won't have that available on the GMAT

b) you can use the formula

nCr = (n!)/[(r!)((n-r)!)]

c) You can use Pascal's Triangle

http://en.wikipedia.org/wiki/Pascal%27s_triangleWriting out to the six row of Pascal's triangle (the top 1 counts as the "zeroth" row), we get

Attachment:

Pascal's triangle, up to n = 6.JPG [ 12.63 KiB | Viewed 885 times ]
You see, nCr is the rth entry on the nth row of Pascal's triangle. (We have to remember to start counting at zero for both the rows and the entries.)

In the six row, the zeroth entry is 1, the first entry is 6, and the second entry is 15, so 6C2 = 15.

Notice the symmetry ----- 6C4 = 6C2, and more generally, nCr = nC(n-r)

When r = 2, a great shortcut you can use is

nC2 = n(n-1)/2

That's very useful for things like 30C2, in which case it would be prohibitive to write out Pascal's triangle all the way.

Does all this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep