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605-655 Level|   Combinations|            
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srijatverma
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Bunuel
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Bunuel,
Could you please explain why do you deduct one?
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Bunuel
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Bunuel,
Could you please explain why do you deduct one?

4C2 is the number of ALL two-element subsets. We subtract one subset {2, 4} to get the desired number.
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srijatverma
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
You can list the subsets in about 30 seconds.

{1,2,3,4}

Two-element subsets that do not contain the pair of elements 2 and 4

Element 1 paired with each possibility
Element 2 paired with each possibility
Element 3 - same

{1,2}
{1,3}
{1,4}
{2,3}
{3,4}

There are 5 subsets that satisfy the conditions of the prompt.

Answer D
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srijatverma
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

The number of 2-element subsets that can be formed from a 4-element set is 4C2 = (4 x 3)/2! = 6. Since the pair of elements 2 and 4 is only 1 of these 6 subsets, we have 6 - 1 = 5 subsets that do not contain the pair of elements.

Answer: D
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Hi All,

You can use the Combination Formula to answer this question, although we have to do a little bit of extra work at the end. Since the number of possible outcomes is so small, you could also list them all out.

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.

Final Answer:

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srijatverma
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

hi, :-)
can anyone explain what does subset mean ? I didn't understand the question ...
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dave13

hi, :-)
can anyone explain what does subset mean ? I didn't understand the question ...

Hey dave13 ,

Subset means part of the set.

Here : Set is {1,2,3,4}

Subset could be {1} , {2}, {1,2},{1,2,3}, etc. where {1} is one element subset, {1,2} is two element subset and so on.

Question is asking how many such two element subsets could be made.

Does that make sense?
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Correct option : 5 - D
Set {1,2,3,4}
Option : {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4} = 6
Not required : {2,4} = 1
Answer :6 - 1 = 5
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Asked: How many two-element subsets of {1,2,3,4} are there that do not contain the pair of elements 2 and 4?

Total two-element subsets of {1,2,3,4} = 4C2 = 6 subsets
Subsets that contain the pair of elements 2 and 4 = {2,4} : 1 subset
Number of two-element subsets of {1,2,3,4} are there that do not contain the pair of elements 2 and 4 = 6-1 = 5

IMO D
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Hey, so in case the question asks for 3 element we can choose it in 4C3 ways = 4 ways
So sets not having 2 and 4 will be 3

123
134
234 (this one will substract)
241

I hope my understanding is correct

Posted from my mobile device
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JenniferMassey
Hey, so in case the question asks for 3 element we can choose it in 4C3 ways = 4 ways
So sets not having 2 and 4 will be 3

123
134
234 (this one will substract)
241

I hope my understanding is correct

Posted from my mobile device

Hi JenniferMassey,

You are correct that there would be 4 possible groups of 3 elements. However, assuming that you want to remove any options that include BOTH a '2' and a '4', you would end up removing TWO of the four options (both 234 and 241).

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com
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Hi,

but the question says 2and 4, not 2 or 4. That would only one be one case for 2 and 4.
ScottTargetTestPrep


The number of 2-element subsets that can be formed from a 4-element set is 4C2 = (4 x 3)/2! = 6. Since the pair of elements 2 and 4 is only 1 of these 6 subsets, we have 6 - 1 = 5 subsets that do not contain the pair of elements.

Answer: D
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kashisshhhhh
Hi,

but the question says 2and 4, not 2 or 4. That would only one be one case for 2 and 4.


It means you only exclude the specific pair {2,4}, not all subsets containing 2 or 4 separately. So every pair is allowed except {2,4}, which leaves 5 valid subsets.
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