2008 AIME II Problems/Problem 12

Problem

There are two distinguishable flagpoles, and there are flags, of which are identical blue flags, and are identical green flags. Let be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when is divided by .

Solution

The well known problem of ordering elements of a string of elements such that none of the elements are next to each other has solutions. (1)

We generalize for blues and greens. Consider a string of elements such that we want to choose the greens such that none of them are next to each other. We would also like to choose a place where we can divide this string into two strings such that the left one represents the first pole, and the right one represents the second pole, in order up the pole according to position on the string.

However, this method does not account for the fact that the first pole could end with a green, and the second pole could start with a green, since the original string assumed that no greens could be consecutive. We solve this problem by introducing an extra blue, such that we choose our divider by choosing one of these blues, and not including that one as a flag on either pole.

From (1), we now have ways to order the string such that no greens are next to each other, and ways to choose the extra blue that will divide the string into the two poles: or orderings in total.

However, we have overcounted the solutions where either pole has no flags, we have to count these separately. This is the same as choosing our extra blue as one of the two ends, and ordering the other blues and greens such that no greens are next to each other: for a total of such orderings.

Thus, we have orderings that satisfy the conditions in the problem: plugging in and , we get .