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 tackleHence 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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