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by Bullzi » Sun Sep 27, 2015 7:14 pm
Hello,
This is how I could approach, I'll let the experts confirm if this approach is right..

Listing the combinations or using 4C2 to pick 2 out of 4,
12 21 31 41
13 23 32 42
14 24 34 43
and canceling out duplicates, we have 6 distinct values of which the only combination that has 2 and 4 is 24 removing which we have 5 values. So, I'll go with Option D

Thanks
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by [email protected] » Sun Sep 27, 2015 9:38 pm
Hi Bullzi,

Your approach is great, although it has some redundant elements to it. If you're going to use the Combination Formula, then you don't have to list out all of the duplicate elements.

4C2 = 4!/(2!2!) = 6 pairs

The pairs would be 12, 13, 14, 23, 24 and 34

Since we're asked to NOT use 24, there are 5 options remaining.

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by Brent@GMATPrepNow » Mon Sep 28, 2015 11:06 am
How many two-element subsets of {1, 2, 3, 4} are there that do not contain the pair of elements 2 and 4?
(A) One
(B) Two
(C) Four
(D) Five
(E) Six
Always check the answer choices before solving a math question on the GMAT.
Given the very small answer choices, students should consider just listing all of the possible subsets.

They are:
(1,2)
(1,3)
(1,4)
(2,3)
(3,4)

The answer is D Five

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by Brent@GMATPrepNow » Mon Sep 28, 2015 11:07 am
Listing and counting is often a useful approach on the GMAT. Here's a free video on this topic: https://www.gmatprepnow.com/module/gmat- ... /video/773

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by gmatbeater1989 » Mon Nov 23, 2015 11:37 am
oquiella wrote:How many two-element subsets of {1,2,3,4} are there that do NOT contain the pair of elements 2 and 4.

A. One
B. Two
C. Four
D. Five
E. Six
4C2 - 1 = 5 (D)
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