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by gmat740 » Tue Jul 28, 2009 12:00 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


OA-C
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by kaulnikhil » Tue Jul 28, 2009 2:52 am
take permutation since order imp
8p3 = 332
now subtract 32 from this as 32 wernt distributed hence
ans = 332 - 32
= 300
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by PussInBoots » Tue Jul 28, 2009 8:51 am
The key word is different letters, hence total # of possobilities is 8 * 7 * 6
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by mmslf75 » Sat Dec 26, 2009 10:13 pm
gmat740 wrote: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


OA-C
8 * 7 * ^ = 336
336 - 36 = 300 simple!!

Does "" If no sequence was assigned to more than one participant "" mean that each got different sequences ??
If so, then I guess my reasoining is right !!
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by valleeny » Sat Dec 26, 2009 10:30 pm
Total possible sequences = 8P3 = 336
Total participants = 336-36=300
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