Stumped by this combinatorics problem

zagcollins

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

sudhir3127 · Posted: Mon Jul 14, 2008 6:55 am Post subject:

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

zagcollins · Posted: Mon Jul 14, 2008 8:29 am Post subject:

thanks..no wonder i didnt get the answer..mistook it for a combination problem....

evansbd · Posted: Thu Jul 17, 2008 7:16 am Post subject: Combinations vs Permuation

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?

parallel_chase · Posted: Thu Jul 17, 2008 7:48 am Post subject: Re: Combinations vs Permuation