Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 3"

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Latest revision as of 16:32, 2 April 2008

Problem

A mother purchases 5 blue plates, 2 red plates, 2 green plates, and 1 orange plate. How many ways are there for her to arrange these plates for dinner around her circular table if she doesn't want the 2 green plates to be adjacent?

Solution

We apply the complement principle: we find the total number of cases in which the 2 green places are adjacent, and subtract from the total number of cases.

There are $\frac{10!}{5!2!2!1!} = 7560$ ways to arrange the plates in a linear fashion. However, since the plates are arranged in a circle, there are $10$ ways to rotate the plates, and so there are $7560/10 = 756$ ways to arrange the plates in a circular fashion (consider, for example, fixing the orange plate at the top of the table).

If the two green plates are adjacent, we may think of them as a single entity, so that there are now $9$ objects to be placed around the table in a circular fashion. Using the same argument, there are $\frac{9!}{5!2!1!1!} = 1512$ ways to arrange the objects in a linear fashion, and $1512/9 = 168$ ways in a circular fashion.

Thus, the answer is $756 - 168 = \boxed{588}$.

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 2
Followed by
Problem 4
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