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by manpsingh87 » Fri Apr 01, 2011 9:08 am
In how many ways can 3 letters can be selected from 3 identical A's,3 identical B's,3 identical C's ?

A)15
B)10
C)27
D)25
E)20
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by jaymw » Fri Apr 01, 2011 9:34 am
For the first letter, there are 3 possibilites: A, B, or C.

For the second and third letter, the same holds true.

Therefore the number of possibilites is 3*3*3=27

C is correct.

What's the source of this? IMO, it does not seem very GMAT-like.
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by GMATGuruNY » Fri Apr 01, 2011 9:39 am
manpsingh87 wrote:In how many ways can 3 letters can be selected from 3 identical A's,3 identical B's,3 identical C's ?

A)15
B)10
C)27
D)25
E)20
All 3 letters the same:
AAA, BBB, CCC = 3 ways.

Two letters the same, the other different:
AAB, AAC = 2
BBA, BBC = 2
CCA, CCB = 2
2+2+2 = 6 ways.

All 3 letters different:
ABC = 1 way.

Total number of ways = 3+6+1 = 10 ways.

The correct answer is B.

The solution above assumes that we want to count the number of different combinations that can be chosen.
The wording of the question is unclear.
Is AAC a different way of selecting the letters from CAA?
If so, then the solution would be 3*3*3 = 27 ways.
An actual GMAT question would use more precise wording.
Last edited by GMATGuruNY on Fri Apr 01, 2011 9:48 am, edited 1 time in total.
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by jaymw » Fri Apr 01, 2011 9:47 am
GMATGuruNY, your explanation makes sense!

But is the word "identical" enough to conclude that the order does not matter in this problem? I do concur that the question does not say anything about building a 3-letter word, tho.
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by manpsingh87 » Fri Apr 01, 2011 10:04 am
Mitch is absolutely correct...!!!
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