Hi - I had a question regarding this problem. I understand that the answer is 3c2 * 3c2 = 9, but the way I originally did it also seems correct to me and I cannot figure out why my method is incorrect. Here was my logic:
- You for sure have to have the distribution center in Chicago, so we just need to figure out how to distribute 4 distribution centers
- Each of the east coast and the west coast needs a center and there are 3 ways to choose each. So, 3 * 3 * 1 so far.
- Finally, the last two centers can be located in any of the cities we haven't used and there are 4 cities still available. So, 4c2, which equals 3.
So, 1 * 3 * 3 * 3 = 27, which isn't even among the answer choices, so I originally picked D. Could anyone point out the flaw in my logic? Thank you.
Combination question
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi Bekerman,
There are a couple of "flaws" in your logic:
The prompt states that there must be 2 centers on the east coast and 2 centers on the west coast...
1) Your "last step" (the last 2 centers can be in ANY of the 4 cities still available) is NOT correct.
2) The "order" of the cities does NOT matter, so however you choose to do the math, you MUST remove any duplicate entries from the total. Your approach seems to revolve around picking a "first city" then doing more math to figure out what the "next city" could be. That is incorrect math for this situation.
As an example, choosing cities A and B is the same as choosing cities B and A, so you can't count that as 2 options. The easiest way to remove the duplicates is with the Combination Formula, which is why 3c2 = 3 is the actual number of options on each coast.
GMAT assassins aren't born, they're made,
Rich
There are a couple of "flaws" in your logic:
The prompt states that there must be 2 centers on the east coast and 2 centers on the west coast...
1) Your "last step" (the last 2 centers can be in ANY of the 4 cities still available) is NOT correct.
2) The "order" of the cities does NOT matter, so however you choose to do the math, you MUST remove any duplicate entries from the total. Your approach seems to revolve around picking a "first city" then doing more math to figure out what the "next city" could be. That is incorrect math for this situation.
As an example, choosing cities A and B is the same as choosing cities B and A, so you can't count that as 2 options. The easiest way to remove the duplicates is with the Combination Formula, which is why 3c2 = 3 is the actual number of options on each coast.
GMAT assassins aren't born, they're made,
Rich
