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Counting Methods

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by Holden123 » Fri Jul 29, 2016 6:49 am
Hi, I am having trouble to understand why a subset of 2 of the set S = {A,B,C,D,E} which equals 5 over 2 = 10 equals 5 over 3. It makes sense mathematically using the formula (n!)/(k!(n-k)!), however I don't get the following:
As per OG, every 2-element subset chosen from a set of 5 elements corresponds to a unique 3-element subset consisting of the elements NOT chosen? (see OG 2016 p.118).
Can someone please explain? Merci.
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by Brent@GMATPrepNow » Fri Jul 29, 2016 7:02 am
Holden123 wrote:Hi, I am having trouble to understand why a subset of 2 of the set S = {A,B,C,D,E} which equals 5 over 2 = 10 equals 5 over 3. It makes sense mathematically using the formula (n!)/(k!(n-k)!), however I don't get the following:
As per OG, every 2-element subset chosen from a set of 5 elements corresponds to a unique 3-element subset consisting of the elements NOT chosen? (see OG 2016 p.118).
Can someone please explain? Merci.
Let's use an easier example.
You have 5 friends, and want to invite 4 of them to attend a party.
Choosing 4 friends to attend is the SAME as choosing 1 friend not to attend.

Another example.
There are 5 people. Your task is to create two teams: team A and team B
Team A will have 3 players and team B will have 2 players.
In how many different ways can we divide the 5 people into the 2 teams?
One approach is to select the 3 players for team A, and the remaining 2 players will be on team B
Another approach, is to select the 2 players for team B, and the remaining 3 players will be on team A

As you might imagine, each approach should have the same number of outcomes.
For the first approach, we must find 5C3 and for the second approach, we must find 5C2
So, it must be the case that 5C3 = 5C2

Does that help?

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by regor60 » Fri Jul 29, 2016 7:03 am
If you pick 2 from the 5 elements, then the formula is 5!/2!3!, as you describe.

If you pick 3 from the 5 elements, then the formula is 5!/3!2!

If you inspect both of these formulas, you can see they are identical, but for the order of multiplication in the denominator, which doesn't matter.

So, picking 2 unique from 5 is the same as picking 3 unique from 5.

This becomes clear if you realize that once you've picked 2 from 5, there are 3 left over, so one can just as easily say the 3 were picked
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by [email protected] » Fri Jul 29, 2016 8:54 am
Hi Holden123,

It looks like both Brent and regor60 have answered your initial question, so I won't rehash any of that here. Instead, I'll point out that this rule applies to all similar calculations that fit this pattern.

re: The number of groups of 4 from a total of 9 is the same as the number of groups of 5 from a total of 9.

9!/4!5! = 9!/5!4!

re: The number of groups of 3 from a total of 10 is the same as the number of groups of 7 from a total of 10.

10!/3!7! = 10!/7!3!

Etc.

While this concept isn't likely to show up on Test Day, you might end up seeing it if you score a high level in the Quant section.

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by Matt@VeritasPrep » Thu Aug 04, 2016 9:42 pm
It might be easier to see like so:

{1, 2, 3, 4, 5}

splits into

{1, 2} {3, 4, 5}

So choosing the set {1, 2} creates the complement {3, 4, 5}, and vice versa, so (5 choose 2) = (5 choose 3).
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