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Comment on Committee of 2 Men & 2 Women
How is 6C2 =15
nCr = n!/[r!(n-r)!]
nCr = n!/[r!(n-r)!]
So, 6C2 = 6!/[2!(6-2)!]
= 6!/[2!4!]
= (6)(5)(4)(3)(2)(1)/(2)(1)(4)(3)(2)(1)
= 15
Here's the video that first introduces combinations: https://www.greenlighttestprep.com/module/gre-counting/video/787
Here's a video that shows how to calculate combinations in your head: https://www.greenlighttestprep.com/module/gre-counting/video/789
I know the outcome of each
We can still use the FCP, BUT
We can still use the FCP, BUT we need to recognize when to use combinations within our solution.
This is what I have done in the video solution.
We took the task of creating the committee, and broke it into STAGES.
STAGE 1: Select the 2 men
STAGE 2: Select the 2 women
Aside: Notice that the outcomes of each stage are different. In one stage, we are selecting men and in the other stage, we are selecting women. So, we can still apply the FCP.
However, when we get to stage 1, we must recognize that IF we try to break that stage into smaller stages (e.g., select 1 man and then another man),the outcomes are the same. In both cases, we're selecting a man to be on the committee, which means we can't use the FCP. Instead, we must use combinations. That is, we can select 2 men from 6 men in 6C2 ways (15 ways). So, we can complete stage 1 in 15 ways.
The same goes for stage 2 (as in the video solution).
NOTE: If we were to break STAGES 1 and 2 into smaller stages, there is a strategy we COULD use to correct out calculations, but I'd rather not get into it, as it will cause more confusion than it will help.
Is this not a restriction
I wouldn't classify questions
I wouldn't classify questions as either restriction questions or non-restriction questions. That will lead to problems.
Also, we can pretty much classify all counting questions as restriction questions. Consider, for example, this question:
"Company X has 8 employees. In how many different ways can we select 3 employees to be on the Party-Planning committee?"
We COULD say that this is a restriction question, because we are restricted to having exactly 3 people on the committee.
Cheers,
Brent
Okay. Makes more sense now,
What is the explanation
As we try to set up the stage
As we try to set up the stage is to use in an FCP solution, we should ask, "Does the outcome of each stage differ from the outcomes of other stages?" (this is addressed in the following video: https://www.greenlighttestprep.com/module/gre-counting/video/788).
The FCP stages for the solution to the above problem would look something like this:
STAGE 1: Select a man to be on the committee
STAGE 2: Select another man to be on the committee
STAGE 3: Select a woman to be on the committee
STAGE 4: Select another woman to be on the committee
Notice that the outcomes for STAGES 1 and 2 are the same. In both cases, a MAN is selected to be on the committee. Since the outcomes are the same, we can't use the FCP.
The same applies to the outcomes for STAGES 3 and 4.
-----------------------------------
To show WHY we can't use the FCP, consider the exact same question, EXCEPT we only have 2 men and 2 women to choose from. This means, we must select all four people to be on the committee. So, there's only 1 possible outcome for this new question.
However, if we try to use the FCP, will get a different answer.
STAGE 1: Select a man to be on the committee: 2 men to choose from, so we can complete STAGE 1 in 2 ways.
STAGE 2: Select another man to be on the committee: We can complete the stage in 1 way.
STAGE 3: Select a woman to be on the committee: 2 women to choose from, so we can complete STAGE 2 in 2 ways.
STAGE 4: Select another woman to be on the committee: We can complete the stage in 1 way.
By the FCP, the answer = (2)(1)(2)(1) = 4, which is clearly incorrect.
Does that help?