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Stumped by this combinatorics problem

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Stumped by this combinatorics problem

by zagcollins » Mon Jul 14, 2008 6:28 am
Each participant in a certain study was assigned a sequence of 3 different letters from 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, for example, that the sequence A, B, C is different from the sequence C, B, A.)

A. 20
B. 92
C. 300
D. 372
E. 476
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by sudhir3127 » Mon Jul 14, 2008 6:55 am
hi the answer is 300.

its a permutation problem as the sequence is important.

arranging 2 alphabets from 8 .. 8P3 ways = 336

36 sequences unassigned therefore its 336-36= 300
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by zagcollins » Mon Jul 14, 2008 8:29 am
thanks..no wonder i didnt get the answer..mistook it for a combination problem....
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Combinations vs Permuation

by evansbd » Thu Jul 17, 2008 7:16 am
Is the only difference between the two is if order matters?

For example, if you know that order matters, you know its a permutation, and if order does not matter then it is a combination?
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Re: Combinations vs Permuation

by parallel_chase » Thu Jul 17, 2008 7:48 am
evansbd wrote:Is the only difference between the two is if order matters?

For example, if you know that order matters, you know its a permutation, and if order does not matter then it is a combination?
Well its true usually. I say usually because a lot of times word "order" is not clearly defined in the question. It is better to understand the concept rather than defining it.
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