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by alex.gellatly » Sat Jul 28, 2012 11:04 pm
Each participant in a certain study was assigned a sequence of 3 different letters form the set {A,B,C,D,E,F,G,H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, A,B,C is different from C,B,A.)

20
92
300
372
476

I got the correct answer, just putting it out there for different approaches. Enjoy.
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by Birottam Dutta » Sat Jul 28, 2012 11:15 pm
Lemme have a crack at it.

Since order matters, it must be permutation.

Now, 3 letters can be selected from 8 in 8P3 ways = 336 ways.

Since 36 possible sequences are not assigned to any participant,

Number of distinct participants = 336-36 = 300.

Hope this is correct!

Answer C!
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by alex.gellatly » Sun Jul 29, 2012 7:25 pm
Birottam Dutta wrote:Lemme have a crack at it.

Since order matters, it must be permutation.

Now, 3 letters can be selected from 8 in 8P3 ways = 336 ways.

Since 36 possible sequences are not assigned to any participant,

Number of distinct participants = 336-36 = 300.

Hope this is correct!

Answer C!
You got it
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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by Lifetron » Mon Jul 30, 2012 3:50 am
8p3-36 = 336-36 = 300
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