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A certain company assigns employees to offices in such a way [#permalink]

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19 Jun 2008, 07:17

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A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices? A. 5 B. 6 C. 7 D. 8 E. 9

Yeah I was thinking the same thing, but I think combinatorics at this level of number of options would be overkill and would take more time than just listing them out. The problem is that you have to arrange all options for each office 2(3C0 + 3C1 + 3C2 + 3C3) THEN you have to subtract all the options that overlap. That just boggles the mind.

Maybe there is an easy combination that I am missing for this one.

Yep you are right. Because we know that figuring out one offices population is all we need with two offices we can use set theory. With 5 offices we have multiple possibilities where a office would be empty. So knowing the sets for one is not enough. Hmm, now I want to figure out that one

Ok this took some back figuring, but I believe a rule for this type of problem holds as follows:

2 offices 3 employees : 2^3 5 offices 8 employees : 5^8 = 390625 So take n office with x employees : n^x

This is just an extension off of a pattern I observed... I was not able to come up with a proof (especially at work ) But what is funny is that the rule turns out to be "Each employee can be assigned in 2 ways - hence in total it is 2x2x2 = 8" which is what iamcartic said. I am not sure how to get at it from set theory, but it does show that set theory and combinatorics are not so dissimilar.