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Re: GRE Math Challenge #41 [#permalink]
sagar wrote:
expert advice is needed.I dont agree with the answer.



The person answering this is expert.

@sandy can you please elaborate how 4C2 is further resolved and in what conditions we use it?
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Re: GRE Math Challenge #41 [#permalink]
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The way we can select 2 objects from a list of 4 is \(C^{4}_{2}=\frac{4!}{2! \times 2!}=6\). Hence there are 6 ways of choosing a pair of number from 4 numbers.

Now when we eyeball the data we see only one number pair whose sum is 8 (2+4).

Hence the probability is \(\frac{1}{6}\)
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Re: GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]
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Option A :

Different Approach

Sum of two Numbers to be 8 which leaves us with only two possible numbers (2 & 6)

So for the 1st Number we have 2 options either to select 2 or 6 from total of 4 numbers : 2/4

For the 2nd Number since we have already selected 1 out of 2 or 6 , the # of ways to select the other number : 1/3 (Since 1 Number is already selected we only have 3 Numbers Left)

Solutions = 2/4*1/3 = 2/12 = 1/6
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Re: GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]
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One more approach:

1) The set consists of {2, 4, 6, 8}
2) Only 2 and 6 add up to 8
3) Two ways to get 8, choose 6 & 2 or 2 & 6
4) Therefore, P = (1/4) * (1/3) * 2 = (2/12) = (1/6)
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Re: GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]
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