2 distributors in east coast and 2 in west coast. And 1 in Midwest.

So,

East Coast= 3C2: \(\frac{3!}{2!1!}\)= 3.

West Coast= 3C2: \(\frac{3!}{2!1!}\)= 3.

Midwest= 1C1= 1

So The total number of ways will be,

\(3*1*3= 9\).

Answer is B.
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1) There are 3 cities on the East Coast; we must put a distribution center in 2 of them.
2) There is 1 city in the Midwest; we must put a distribution center there.
3) There are 3 cities on the West Coast; we must put a distribution center in 2 of them.

Since the "order" of the distribution centers does NOT matter, we're dealing with a Combinations situation.

East Coast: 3c2 = 3!/[2!1!] = 3 ways to place the distribution centers
Midwest: 1c1 = 1 way to place the distribution center
West Coast: 3c2 = 3!/[2!1!] = 3 ways to place the distribution centers

We have to multiply these results to calculate the TOTAL possible combinations of distribution centers:

(3)(1)(3) = 9 different ways to place the distribution centers.

in west coast 2out of 3
in east coast 2out of 3
and mid west 1 of 1..
total ways 3c2*3c2*1c1=9.. ans B
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Cant we have 3 on east coast and only one on the west.

Hi,
NO, we have to choose two in each coast since it is given the coast..
..

had the wordings been ..
If the company must add one distribution centers in the Midwest, and only one center can be added in each city, in how many ways can the management allocate the distribution centers?
in that case the WAY mentioned by you will also come into play..

Hope it helped
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Hi chetan2u

can you elaborate? I have a similar doubt. We need to pick two centers in each coast as per the question.
I think as below
1st distribution center in east coast can have 3 choices
2nd distribution center in east coast can have 2 choices
Similar case with west coast.
Whats wrong with my thinking?

I solved this question the same way:

East Coast : 3x2 (and) West Coast: 3x2 (and) Midwest 1 = 36

Please do point out the error in this logic.

Regards,
Arvind.

This way you are considering that the order of the cities in each coast matters, when it actually doesn't.

It doesn't matter if your picks are:

NY Boston Chicago LA SF

or

Boston NY Chicago LA SF

The cities picked are still the same.

In east coast 3C2 ways..................1
In west coast 3C2 ways..................2
In midwest 1C1 ways ...................3
total= 1*2*3
=9
hence B

Answer is B.. Get is 1:24 should be less though
East coast*Mid coast*west coast
3c2*1c1*3c2
3*1*3
9
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There are 5 Centres and 7 Cities ( EC:3, WC:3, MW:1).
Constraints: 2 Centres affixed to EC; 2 Centres affixed to WC ; 1 Centre affixed to MW.
Ways to Distribute Centres: MW- 1 ; EC: 3C2 ; WC: 3C2.
Hence, Total ways= MW x EC x WC = 1 × 3 × 3 = 9.
Correct Choice: B

We don't necessarily need to factor in the midwest city since there is only one possibility for it; now, actually, the most logical way to think of the problem is that there are three cities that will have or rather choose 2 cities among the 4 cities remaining instead of thinking 4 cities choose 2 spots-

3c2 x 3c2 = 9

The management allocate the distribution centers in : 3C2 * 1C1 * 3C2 = 9

the answer is option B.

Bunuel : please provide the OA after review.

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The OA is B.
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Hi Nevernevergiveup,

If i understood correctly, you would also need to divide by 2! in order to account for any false counts i.e. duplicates. So you would end with 3 such cases per coast.

Hi ALL,

While this question IS wordy, it's really just a Combination Formula question (albeit with a lot of little pieces of information).

By reorganizing the information in the prompt, we are told the following:

1) There are 3 cities on the East Coast; we must put a distribution center in 2 of them.
2) There is 1 city in the Midwest; we must put a distribution center there.
3) There are 3 cities on the West Coast; we must put a distribution center in 2 of them.

Since the "order" of the distribution centers does NOT matter, we're dealing with a Combinatorics situation.

East Coast: 3c2 = 3!/[2!1!] = 3 ways to place the distribution centers
Midwest: 1c1 = 1 way to place the distribution center
West Coast: 3c2 = 3!/[2!1!] = 3 ways to place the distribution centers

We have to multiply these results to calculate the TOTAL possible combinations of distribution centers:

(3)(1)(3) = 9 different ways to place the distribution centers.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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