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# From the set of 6 letters A, B C, D, E and F, there are 20

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Manager
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From the set of 6 letters A, B C, D, E and F, there are 20 [#permalink]

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07 Dec 2003, 04:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

From the set of 6 letters A, B C, D, E and F, there are 20 different 3-letter
subsets that could be selected.

What is the number of 3-letter subsets that include the letter F?

I am wondering if someone got a simpler way..

4x6=24

ABF
ACF
AEF
BAF
BCF
BDF
BEF
CEF
....
....
GMAT Instructor
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Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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07 Dec 2003, 05:31
tzec76 wrote:
From the set of 6 letters A, B C, D, E and F, there are 20 different 3-letter
subsets that could be selected.

What is the number of 3-letter subsets that include the letter F?

I am wondering if someone got a simpler way..

4x6=24

ABF
ACF
AEF
BAF
BCF
BDF
BEF
CEF
....
....

First thing you must determine is "are the subsets ordered?". You should be able to tell by the premise that there are only 20 subsets of 3 in 6. If there are NOT ordered, then you are double counting in your analysis. In any case, there is a much easier way. If you know that something must be including in your subset, perhaps the number of ways the OTHER members can be in the group will yield the correct result....
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Director
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07 Dec 2003, 05:38
Simply take F and you need to select 2 more letters out of 5( without F), which could be done in 10 ways , so you have 10 subsets with F
Manager
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07 Dec 2003, 06:41

The answers are confirmed: 10, and as follows.

ABF
ACF
AEF
BCF
BDF
BEF
CDF
CEF
DEF
SVP
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09 Dec 2003, 03:29
F is taken. How many ways? The only one. Other two letter: 5C2.
So. 1*10=10
SVP
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15 Dec 2003, 16:58
Why ABF is same as BAF. Shouldn't they be different.
I understand thet 6C3 gives 20. Similarly other two letters from the set of 5 can be choosen in 5C2 ways = 10.
If you consider ABF to be same as BAF then this answer is correct.Otherwise lot many combinations are possible.
I hope some can explain why BAF is same as ABF. The question does not mention anything about mere presence of the letter without any order.

Thanks,
Anand.
15 Dec 2003, 16:58
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