Devon wrote:
Thanks for the response everyone. I guess I have trouble sometimes identifying how to go about solving a problem. For example, in
MGMAT V5, p.50 they have the following problem:
A local card club will send 3 representatives to the national conference. If the local club has 8 members, how many different groups of representatives could the club send?
The book makes 8 slots, one for each representative, and then solves with filling the slots out with Y/N, etc.
My approach would be to make three slots, one for each representative sent, which would give 8*7*6=336 possibilities with repetitions. Now, in order to get to the correct answer I would need to divide by 3! (336/6=56). But I don't understand the reasoning behind dividing by 3!. I know that I need to divide to eliminate repetition, but where does the 3! come from in this instance? Can I even solve the problem in this manner?
Further, in their example the answer is 8!/3!5!. It says you need to divide by 3!5! to account for repeats, which in theory makes sense to me, but I don't understand why you're grouping those being sent and those not being sent as "identical" to give you the 3! and 5!. Is it just a rule of combinatorics, in a problem with less slots than options like this, to divide by factorials for BOTH groups (i.e. those going AND those not going)? How would this translate into how I tried to go about solving the problem?
Devon,
You have to be very careful in thinking through the FCP in these situations.
There are 8 members, say {A, B, C, D, E, F, G, H}, and we are going to pick three at random to send to the national conference.
As you say, and the book says, we have eight slots --- 8 for the first slot, 7 for the second, 6 for the third, so 8*7*6 (BTW NEVER NEVER NEVER multiply a number like this out until all canceling it done! Leave it in un-multiplied product form! You are always making life harder than it needs to be if you wind up having to do math with a three-digit number such as 336.)
Now, why isn't 8*7*6 the answer to the question? Because, among the three selected, order doesn't matter --- let's say that D & F & G were the three selected, well, the choices
DFG
DGF
FDG
FGD
GDF
GFD
would count as six different "selections" in the 8*7*6 way of counting, but obviously, those are re-arrangements of the same three people. When we pick a group of n, we don't want to count rearrangements (i.e. permutations) of that group as different choices, so we have to divide by n! --- here, we divide by 3! = 6
The virtue of not multiplying out ---- (8*7*6)/6 = 8*7 = 56 ---- the only math we have to do is single digit math. If you do more than that on the GMAT Quant section, you are working too hard.
Now, let's think about the expression \(\frac{8!}{(3!5!)}\) ----- think of this as \(\frac{8!}{(5!)}*\frac{1}{3!}\) ---- the first piece is the piece that cancels out the 5 people who
won't be selected from the numerator --- the left fraction winds up as (8*7*6), and the right fraction cancels the permutations among the people who
are chosen --- we wind up dividing by (3!), exactly as we had to do in the FCP approach, above.
It's true there's a magic symmetric to the formula nCr = \(\frac{n!}{r!(n-r)!}\), such that the number of combinations of 3 chosen from 8 has to be equal as the number of combination of 5 chosen from 8. This plays into the symmetry of Pascal's Triangle, about which you can read here:
http://magoosh.com/gmat/2012/gmat-math- ... binations/Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)