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# Where am I going wrong

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Manager
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Where am I going wrong [#permalink]

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28 Sep 2010, 09:34
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Renee, has a bag of 6 candies, 4 are sweet 2 are sour. if Jack picks two candies simultaneously how many different combinations can he make

I know the total combinations its just 6!/2!4! from the combination formula. But when I enumerate them I only get XX,XY,YY
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 09:39
Jinglander wrote:
Renee, has a bag of 6 candies, 4 are sweet 2 are sour. if Jack picks two candies simultaneously how many different combinations can he make

I know the total combinations its just 6!/2!4! from the combination formula. But when I enumerate them I only get XX,XY,YY

There is a piece of critical information you are missing here, which is whether all the sweet candies are identical and whether all the sour candies are identical.

If they are identical, then there are only the 3 ways above.

If they are not identical, then the answer is C(6,2) or 15. The difference in this case is that (sweet_1,sweet_2) and (sweet_1,sweet_3) are different combinations and hence counted seperately.
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 09:48
What would be the formula for the situation a gave above when both sweet and sours are identical
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 10:01
Jinglander wrote:
What would be the formula for the situation a gave above when both sweet and sours are identical

You will have to use the cases.

1. Both are identical.

so total number of combinations = 2.....as you can pick the pair either from sweet or sour set.

2. Both are different.

Since there are only two different set of candles, only combination is there.

Thus total = 2+1 = 3.

Suppose you have to calculate the number of possible 4 letter words with or without meaning from MATHEMATICS. You will again have to build the cases.
Eg. all different, 2 same 2 different, 2 pair of same letters.
This question is a perfect example of combination of permutation & combination.
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 10:05
You seem to be enumerating the answer since the set is small. But can we get to a formula that could in other cases such as 10 different catagories of candies in which we choose two.
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 10:12
Jinglander wrote:
What would be the formula for the situation a gave above when both sweet and sours are identical

Surprisingly enough, dealing with identical objects is a much harder combinatorial problem than non-identical objects. For the sake of mathematical curiosity (if there is such a thing) I can show you how to do this, but please don't get confused by this, much of this technique would never be tested on the GMAT :

In this question we know that each set can supply 0,1 or 2 objects
And the total objects to be selected is also 2.

Now let the polynomial (1+x+x^2) represent the objects selected from set 1 (sweet). Here 1=x^0 represents the case 0 objects are selected, x that 1 is selected and x^2 that 2 are selected.
Similarly for the set sour, there is polynomial (1+x+x^2)

Now the idea is that if I take the product $$(1+x+x^2)_{sweet} * (1+x+x^2)_{sour}$$ then the coefficient of the term x^2 in this product represents the number of ways to pick 2 objects

Explanation : To form x^2, I will have to pick x^a from set 1 and x^b from set 2 such that a+b=2. And also note there is no distinction between the a's and b's, so within set 1 and within set 2, all is identical. This is exactly the combinatorial problem I am trying to tackle

Hence the solution is the coefficient of x^2 in (1+x+x^2)^2

If you expand this out (formula in identities section) you will notice the coefficient of x^2 is 3. And this is our answer.

All this is a lot of effort for little reward it seems, but this technique is priceless, if the number of objects you are dealing with is larger, then manual counting is not an option

Hope you find this useful !
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Re: Where am I going wrong [#permalink]

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28 Sep 2010, 10:14
You might also find these notes helpful which deal with a particular kind of identical object problem which is much simpler when the number of objects to be chosen is unrestricted (How many ways to pick 0 or more sweets from 4 identical sweet and 2 identical sour ones)
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Re: Where am I going wrong   [#permalink] 28 Sep 2010, 10:14
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