Hi, there. I'm happy to help with this.

This problem involves the Fundamental Principle of Counting. You may find these blogs helpful for background.
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/
http://magoosh.com/gmat/2012/gmat-permu ... binations/

First of all, one thing that will make a huge difference is who the lead singer is. We have 30 choices for the lead singer. So count that first. That's where the 30 comes from.

Now, for each choice of lead singer, there are 29 folks remaining. This means, for a choir of 28 counting the lead singer, we need to select 27 of these remaining 27 people. Mathematically, that's

29C27

That would be ugly to calculate --- instead, we'll use the symmetry of Pascal's Triangle to reduce that to 29C27 = 29C2 --- another way to say that is --- choosing the 27 of the 29 are included is entirely equivalent to choosing the two of 29 who will not be excluded and told to go home. That's also why 29C27 = 29C2.

Here, it's useful to know the handy shortcut:

nC2 = n(n-1)/2 (which, incidentally, is also the sum of the first n integers, but that's another story.)

So,

29C2 = (29*28)/2 = 29*14

For each one of the 30 choices for lead singer, we could form 29*14 different choirs to back up that lead singer, for a grand total of

30*29*14

Does all this make sense? That was a lot. Please let me know if you have further questions on any of this.

Mike
_________________

Mike McGarry
Magoosh Test Prep

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_triangle

Writing out to the six row of Pascal's triangle (the top 1 counts as the "zeroth" row), we get

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