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Combination with repetition and restrictions

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Combination with repetition and restrictions  [#permalink]

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New post 05 May 2011, 06:18
Here is the question:

There is a bowl that contains one apple, two bananas, and four cherries. How many unique groups of four fruit can be picked.

I can easily find the answer by manually counting, but I cannot think of a systematic approach to apply to any similar question.

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Re: Combination with repetition and restrictions  [#permalink]

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New post 05 May 2011, 07:54
1
considering that repetitions aren't allowed -

groups of cherries to be chosen = 4c4
groups of cherries and apples to be chosen = 4c2 * 2c2
groups of cherries and banana to be chosen = 4c3 * 1c1

adding all we get 11.
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New post 05 May 2011, 08:12
amit2k9 wrote:
considering that repetitions aren't allowed -

groups of cherries to be chosen = 4c4
groups of cherries and apples to be chosen = 4c2 * 2c2
groups of cherries and banana to be chosen = 4c3 * 1c1

adding all we get 11.


I think you are incorrect.

Just Cherries: 4c4 = 1
Cherries and apple: there is only one apple, so that is forced. It would be 1c1*3c3=1
Cherries and Banana: there are only two bananas, so it can either be 1 banana and 3 cherries, or 2 bananas 2 cherries, so we are really just deciding how many bananas (1 or 2) and the cherries are forced, so that would be 2
Finally a group of all three must include, 1 apple, an either 1 or 2 bananas, cherries are forced by the amount of bananas - that is 2 as well.

Total is 6

I am looking for a general formula, assuming I need to pick k combinations from n items consisting of u unique items (u1, u2, u3...) with maximums on each unique item m (m1, m2, m3...) how can I write such a general formula for this kind of question?
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New post 05 May 2011, 08:40
Ok here goes a potential solution

Let k represent the number of items we are going to pick, in this case 4.
let u represent the number of unique items, in this case 3

Assuming there were infinite amount of unique items, we can calculate the combination as: (k+u-1)C(k)

Now we have to subtract the cases where we violate our maximums.
for every unique item, we have a maximum we will call m, the number of violations of that maximum would be (u+k-m-2)C(k-m-1) because I force a violation of the maximum and then determine how many violations there are.

in our case, k=4, u=3, m1=1 and m2=2 (cherries are infinite for all intents and purposes because they are >=k)

Without limitations, it would be (k+u-1)C(k) = 6C4 = 15
subtract (k+u-m1-2)C(k-m1-1) = 4C2 = 6
and (k+u-m2-2)C(k-m2-1)= 3C1 = 3
so 15-6-3 = 6

Does this make sense? Is it correct?
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Re: Combination with repetition and restrictions  [#permalink]

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New post 05 May 2011, 09:11
randompattern wrote:
Here is the question:

There is a bowl that contains one apple, two bananas, and four cherries. How many unique groups of four fruit can be picked.

I can easily find the answer by manually counting, but I cannot think of a systematic approach to apply to any similar question.


Why not \(C^{7}_{4}=35\)? All groups of 4 fruits will be unique.
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New post 05 May 2011, 11:25
fluke wrote:

Why not \(C^{7}_{4}=35\)? All groups of 4 fruits will be unique.


Because the four cherries are identical and the two bananas are identical. Your method grossly over counts.
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Re: Combination with repetition and restrictions  [#permalink]

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New post 05 May 2011, 11:46
randompattern wrote:
fluke wrote:

Why not \(C^{7}_{4}=35\)? All groups of 4 fruits will be unique.


Because the four cherries are identical and the two bananas are identical. Your method grossly over counts.


I see;
CABB
CCBB
CCAB
CCCA
CCCB
CCCC

Just 6.

Wonder how to formulate that. I'll just skip.
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Re: Combination with repetition and restrictions  [#permalink]

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New post 10 May 2011, 19:45
randompattern wrote:
Ok here goes a potential solution

Let k represent the number of items we are going to pick, in this case 4.
let u represent the number of unique items, in this case 3

Assuming there were infinite amount of unique items, we can calculate the combination as: (k+u-1)C(k)

Now we have to subtract the cases where we violate our maximums.
for every unique item, we have a maximum we will call m, the number of violations of that maximum would be (u+k-m-2)C(k-m-1) because I force a violation of the maximum and then determine how many violations there are.

in our case, k=4, u=3, m1=1 and m2=2 (cherries are infinite for all intents and purposes because they are >=k)

Without limitations, it would be (k+u-1)C(k) = 6C4 = 15
subtract (k+u-m1-2)C(k-m1-1) = 4C2 = 6
and (k+u-m2-2)C(k-m2-1)= 3C1 = 3
so 15-6-3 = 6

Does this make sense? Is it correct?


Won't this method be specific to this problem, because we are considering that cherries are infinite for all intents and purposes.
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New post 11 May 2011, 06:01
Good Point Varun,

An additional problem with my approach is that I failed to consider the possibility of violating two maximums at once. In my case, it is impossible, but in a general case it would be possible, and the method I provided would over count maximum violations and end up with a lower final answer.

Furthermore bec. of your point of regarding the cherries as infinite for all intents and purposes, and because if you take all apples and all bananas you will have less than four fruit, In this specific case we can simplify a heck of a lot. We can pretty much ignore the cherries and just look at the unique combinations of amounts of apples and bananas.

If a=1, there are a+1 possible amounts (0 or 1)
and if b=2, there are b+2 possible amounts (0,1,2)
as we mentioned before, the amount of cherries will be automatic based on the amount of apples and bananas, so it can be ignored.

so it's like having a menu with 2 appetizers and 3 entrées. You have 2*3=6 possibilities.
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Re: Combination with repetition and restrictions  [#permalink]

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New post 11 May 2011, 10:19
I think the best approach to solve this type of question is by formulation a solution considering all the scenarios rather than trying to find a generalized solution.

Moreover if the question is too tricky and demanding lot of effort, it won't be worth on the D-Day!! :-D

PS: Thats solely my view!! :-D :-D
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Re: Combination with repetition and restrictions  [#permalink]

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New post 11 May 2011, 19:08
randompattern wrote:
Here is the question:

There is a bowl that contains one apple, two bananas, and four cherries. How many unique groups of four fruit can be picked.

I can easily find the answer by manually counting, but I cannot think of a systematic approach to apply to any similar question.


If facing such a question in GMAT, I would suggest counting in an orderly manner.
4 cherries - 1 way
3 cherries - 2 ways (With a banana or with an apple)
2 cherries - 2 ways (with a banana and an apple or with 2 bananas)
1 cherry - 1 way (We need both bananas and the apple)
0 cherries - not possible

Hence total 6 ways.
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Re: Combination with repetition and restrictions  [#permalink]

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New post 11 May 2011, 19:33
VeritasPrepKarishma wrote:
randompattern wrote:
Here is the question:

There is a bowl that contains one apple, two bananas, and four cherries. How many unique groups of four fruit can be picked.

I can easily find the answer by manually counting, but I cannot think of a systematic approach to apply to any similar question.


If facing such a question in GMAT, I would suggest counting in an orderly manner.
4 cherries - 1 way
3 cherries - 2 ways (With a banana or with an apple)
2 cherries - 2 ways (with a banana and an apple or with 2 bananas)
1 cherry - 1 way (We need both bananas and the apple)
0 cherries - not possible

Hence total 6 ways.


Karishma, Thanks for the solution. This method works if your are solving the question manually. And will be cumbersome if number of items increases. Is there a systematic approach or formula which can be used to solve these type of question.
Any help will be great!! :)
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Re: Combination with repetition and restrictions  [#permalink]

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New post 12 May 2011, 20:08
varunmaheshwari wrote:
VeritasPrepKarishma wrote:
randompattern wrote:
Here is the question:

There is a bowl that contains one apple, two bananas, and four cherries. How many unique groups of four fruit can be picked.

I can easily find the answer by manually counting, but I cannot think of a systematic approach to apply to any similar question.


If facing such a question in GMAT, I would suggest counting in an orderly manner.
4 cherries - 1 way
3 cherries - 2 ways (With a banana or with an apple)
2 cherries - 2 ways (with a banana and an apple or with 2 bananas)
1 cherry - 1 way (We need both bananas and the apple)
0 cherries - not possible

Hence total 6 ways.


Karishma, Thanks for the solution. This method works if your are solving the question manually. And will be cumbersome if number of items increases. Is there a systematic approach or formula which can be used to solve these type of question.
Any help will be great!! :)


I cannot think of any formula that can be used directly here to get the answer (that is not to say that there isn't any, just that there isn't one that is used often). GMAT does not expect you to remember anything other than the basic formulas hence I wouldn't pay much attention to them anyway. Also, GMAT does not give you cumbersome calculations or insipid questions. This question is mainly for practice and to understand the concept of combination used here.

--== Message from the GMAT Club Team ==--

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This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: Combination with repetition and restrictions  [#permalink]

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Re: Combination with repetition and restrictions   [#permalink] 23 Nov 2019, 05:37
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