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Good Perm n Comb Question.

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by mav800rick » Tue May 27, 2008 4:09 pm
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

what's the best way to approach this?
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by netigen » Tue May 27, 2008 7:58 pm
We are trying to choose 3 alphabets out of given 8 and the order of sequencing matters so it's a permutation question.

8P3 = 8 x 7 x 6 = 336

We know that 36 permutations are not used hence the number of participants should be 336-36 = 300
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by ksh » Tue May 27, 2008 10:48 pm
yeah it's pretty straight question

The ans= 8P3-36=300

ans C
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by calande » Wed May 28, 2008 12:52 am
Yep, if you would like more details, Maverick, do not hesitate to ask ;)
Calande.
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I think the answer is A - 20

by caramel3536 » Wed May 28, 2008 12:22 pm
Because the order doesn't matter, you need to divide (just think about it conceptually, that's how I learned it in my class)
so the answer is
8 x 7 x 6
divided by
3!
= 56 total sequences. 56 -36 = 20

What was the answer in the book?
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by chidcguy » Wed May 28, 2008 3:52 pm
The order clearly matters. The Q says that ABC is different from BCA.

My answer is 8 C 3 X 3! - 36 = 300
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