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Combination question

by vinni.k » Sun Jun 12, 2011 9:52 am
Hi,

The following question is a combination question and i was able to solve it, but in the end my method differed from the book's method to get the answer. Please check my method. Below, I have also mentioned the book's method.

A retail company needs to set up 5 additional distribution centers that can be located in three cities on the east coast (Boston, New York, and Washington D.C), once city in the mid-west (Chicago), and three cities on the west coast (Seattle, San Francisco, and Los Angeles). If the company must have 2 distribution centers on each coast and 1 in the mid-west, and only one center can be added in each city, in how many ways can the management allocate the distribution centers ?

(A) 3
(B) 9
(C) 18
(D) 20
(E) 36

Answer is B

My method:-

East coast: 2 slots and 3 candidates (Boston, New York, and D.C) = 3C2 = 3
Mid-west: 1 slot and 1 candidate = 1C1 = 1
West coast: 2 slots and 3 candidates (Seattle, San Francisco, and L.A) = 3C2 = 3

3 + 3 + 1 + 2(and only one center can be added in each city)
= 9 Answer

The difference in the book is that combinations are multiplied:-
3 * 1 * 3 = 9

May be both are correct, but i am not sure. Please help me in figuring this out.

Regards
Vinni
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by Frankenstein » Sun Jun 12, 2011 10:04 am
Hi,
Sorry to say this, but your method is wrong. The 3 coasts are independent. So, choosing from a coast will not have any impact on choosing the other. So, they should be multiplied.
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by finites » Sun Jun 12, 2011 10:54 am
vinni.k wrote:Hi,

The following question is a combination question and i was able to solve it, but in the end my method differed from the book's method to get the answer. Please check my method. Below, I have also mentioned the book's method.

A retail company needs to set up 5 additional distribution centers that can be located in three cities on the east coast (Boston, New York, and Washington D.C), once city in the mid-west (Chicago), and three cities on the west coast (Seattle, San Francisco, and Los Angeles). If the company must have 2 distribution centers on each coast and 1 in the mid-west, and only one center can be added in each city, in how many ways can the management allocate the distribution centers ?

(A) 3
(B) 9
(C) 18
(D) 20
(E) 36

Answer is B

My method:-

East coast: 2 slots and 3 candidates (Boston, New York, and D.C) = 3C2 = 3
Mid-west: 1 slot and 1 candidate = 1C1 = 1
West coast: 2 slots and 3 candidates (Seattle, San Francisco, and L.A) = 3C2 = 3

3 + 3 + 1 + 2(and only one center can be added in each city)
= 9 Answer

The difference in the book is that combinations are multiplied:-
3 * 1 * 3 = 9

May be both are correct, but i am not sure. Please help me in figuring this out.

Regards
Vinni
Buddy in your method how did you get that 2 to add in the end? If you figure that out you will
realise that the method is wrong. Also Frank has given the reasoning below.
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by vinni.k » Mon Jun 13, 2011 2:34 am
finites wrote:
Buddy in your method how did you get that 2 to add in the end? If you figure that out you will
realise that the method is wrong. Also Frank has given the reasoning below.
To be very honest, i know my method is wrong and i also know that "AND" means multiply. But i still don't understand what the last line of the if-statement really means.
If the company must have 2 distribution centers on each coast and 1 in the mid-west, and only one center can be added in each city
From the above statement, i concluded 3C2 for east coast (1 city is still left), 3C2 for west coast (again one city is left), and 1C1 for mid-west.

And only one center can be added in each city --> Does it mean adding one center in the remaining cities of east coast and west coast ?

Experts can you please help ?

Thanks & Regards
Vinni
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by Roy@MasterGmat » Mon Jun 13, 2011 3:50 am
To be very honest, i know my method is wrong and i also know that "AND" means multiply. But i still don't understand what the last line of the if-statement really means.
By 'the last line of the if-statement' I'm assuming you are referring to "and only one center can be added in each city". This means that the management can't allocate 2 or more centers to the same city, i.e., every city may only receive one additional center. This is why it isn't possible to place both west-coast centers in Seattle, for example. In technical terms, this sentence implies the combinations don't allow repetition - every option may only be chosen once.

You're right about the 3C2 for each coast and 1C1 for the mid-west, but the next step is to multiply the resulting numbers, not add them, and the '2' you added in the end is unnecessary. Think of this as a combination of combinations: you have 3 ways to arrange the west coast centers, 3 ways to arrange the east coast centers, and 1 way to arrange the mid-west centers. Thus, arranging all of the new centers demands choosing a west coast arrangement (out of three options) and an east-coast arrangement (out of 3 options): 3*3=9. It's just like choosing a shirt&pants outfit from a wardrobe containing 3 pairs of pants and 3 shirts.
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by GMATGuruNY » Tue Jun 14, 2011 11:01 am
vinni.k wrote:Hi,

The following question is a combination question and i was able to solve it, but in the end my method differed from the book's method to get the answer. Please check my method. Below, I have also mentioned the book's method.

A retail company needs to set up 5 additional distribution centers that can be located in three cities on the east coast (Boston, New York, and Washington D.C), once city in the mid-west (Chicago), and three cities on the west coast (Seattle, San Francisco, and Los Angeles). If the company must have 2 distribution centers on each coast and 1 in the mid-west, and only one center can be added in each city, in how many ways can the management allocate the distribution centers ?

(A) 3
(B) 9
(C) 18
(D) 20
(E) 36

Answer is B
I think of this as a "bucket" problem.
We have 3 buckets: a bucket for the west coast, a bucket for the midwest, and a bucket for the east coast.
To combine from multiple buckets:

1. Determine how many choices can be made from each bucket.
2. Multiply the results.

West coast bucket:
Number of choices = 3C2 = 3.

Midwest bucket:
Number of choices = 1.

East coast bucket:
Number of choices = 3C2 = 3.

Multiplying our choices for each bucket:
3*1*3 = 9.

The correct answer is B.
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by vinni.k » Wed Jun 15, 2011 6:16 am
Roy@MasterGmat wrote:
By 'the last line of the if-statement' I'm assuming you are referring to "and only one center can be added in each city". This means that the management can't allocate 2 or more centers to the same city, i.e., every city may only receive one additional center. This is why it isn't possible to place both west-coast centers in Seattle, for example. In technical terms, this sentence implies the combinations don't allow repetition - every option may only be chosen once.
Thanks Roy for clearing my doubt. I was missing something when I first attempted this question but somehow got the answer. The wording confused me. However, thank you very much.

Thanks Mitch, bucket strategy is quite fascinating.
GMATGuruNY wrote: I think of this as a "bucket" problem.
We have 3 buckets: a bucket for the west coast, a bucket for the midwest, and a bucket for the east coast.
To combine from multiple buckets:

1. Determine how many choices can be made from each bucket.
2. Multiply the results.

West coast bucket:
Number of choices = 3C2 = 3.

Midwest bucket:
Number of choices = 1.

East coast bucket:
Number of choices = 3C2 = 3.

Multiplying our choices for each bucket:
3*1*3 = 9.

The correct answer is B.
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by vikram4689 » Wed Jun 15, 2011 6:37 am
@ EXPERTS : How would the solution change if restriction of no. of centers is not mentioned.

Now each city can have 2 centers (AA,BB,CC,AB,BC,CA). I can solve the problem using the counting method but i would like to know how to solve using P & C.

Answer would be : 6+1+6= 17
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by Frankenstein » Wed Jun 15, 2011 6:58 am
vikram4689 wrote:@ EXPERTS : How would the solution change if restriction of no. of centers is not mentioned.

Now each city can have 2 centers (AA,BB,CC,AB,BC,CA). I can solve the problem using the counting method but i would like to know how to solve using P & C.

Answer would be : 6+1+6= 17
Hi,
I think it should be 6*1*6 = 36.
Cheers!

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by vikram4689 » Wed Jun 15, 2011 7:13 am
oh yes ;) any idea on how to use P & C.
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by Roy@MasterGmat » Wed Jun 15, 2011 7:14 am
Again with the adding instead of multiplying ... ;)

Yes - it's 6*1*6 = 36.

As for a proper calculation, don't bother - stick to counting. Combinations with repetition are not covered by the standard combinations/permutations formulas, so you'll never see a GMAT problem involving them which isn't countable.

To be exact, combinations with repetition may appear, but only when order matters (in n^k situations like 'how many different results are possible when rolling 2 dice?'). In our case, order doesn't matter - 'a center in A and a center in B' is the same as 'a center in B and a center in A'.

Your best bet here is either counting the combinations for each region from scratch, or calculating 3C2 and adding the 3 new "double" options (both in A, both in B, both in C).
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by Frankenstein » Wed Jun 15, 2011 7:24 am
vikram4689 wrote:oh yes ;) any idea on how to use P & C.
Hi,
Okay.. let's say there are n cities and each city has 2 centers.
First center has n cities to pick from: So, nC1 = n
Second center has n cities to pick from: S, nC1 = n
So, total number of cases is n^2
But, as we know except AA,BB,... all other cities are counted twice.
AA,BB,... are n pairs. So, we add this n to n^2 and divide the sum by 2.
So, the required number is (n^2+n)/2 = n(n+1)/2.
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by winniethepooh » Wed Jun 15, 2011 7:59 am
Roy @ Master Gmat wrote:
As for a proper calculation, don't bother - stick to counting. Combinations with repetition are not covered by the standard combinations/permutations formulas, so you'll never see a GMAT problem involving them which isn't countable.

Roy, does this mean that unless the problem explicitly says that there can be 2 centres in one city we should not assume that there can be 2 centres in one city?
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by Roy@MasterGmat » Wed Jun 15, 2011 9:28 am
Roy, does this mean that unless the problem explicitly says that there can be 2 centres in one city we should not assume that there can be 2 centres in one city?
That depends on the exact phrasing, but in this case - no. In itself, the phrasing "5 additional distribution centers that can be located in three cities on the east coast (Boston, New York, and Washington D.C)..." implies that more than one center may be placed in one city. The part limiting the combinations to no-repetition comes a couple of sentences later : "and only one center can be added in each city".
Okay.. let's say there are n cities and each city has 2 centers.
First center has n cities to pick from: So, nC1 = n
Second center has n cities to pick from: S, nC1 = n
So, total number of cases is n^2
But, as we know except AA,BB,... all other cities are counted twice.
AA,BB,... are n pairs. So, we add this n to n^2 and divide the sum by 2.
So, the required number is (n^2+n)/2 = n(n+1)/2.
Interesting technique, nice! You can even take it a little further by describing it in terms of C: it's equal to 3C2+3, which is the number of unique 2 city combinations plus the number of "doubles". Placing 2 centers in n cities with repetition allowed would thus equal nC2+n. This only works for 2 centers; more centers means more than n "doubles", so generalizing this is pretty complicated. Please keep in mind that this is just for fun and completely off-topic - these questions should just be counted.
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by winniethepooh » Wed Jun 15, 2011 9:43 am
Thanks Ron for all your help. I really appreciate it!
I have one more question: If the question is silent about there being more than one centers in one city. As in, what if the question in this question ends like this:
If the company must have 2 distribution centers on each coast and 1 in the mid-west,in how many ways can the management allocate the distribution centers ?
Then, how do we calculate the answer?
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