# 2019 Mock AMC 10B Problems/Problem 24

Let's label the people from three schools with A,B,C. We can consider the case in which A is in seat number 1. Now, we consider to space between two successive A's. There are totally 8 spaces, which can be broken into sum of four positive integers:

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For case of , there are four possible orders. For each order, we can arrange 5 people in the space with 5 seats as or , and then arrange two B's or two C's in two of three remaining spaces, making a total of .

For case of , there are twelve possible orders, , , , , , , , , , , , . For each order, we can arrange 4 people in the space with 4 seats as or , and then arrange the space with two seats as or , and then choose 1 space for and one space for , making a total of .

For case of , there are twelve possible orders, similar to the case above. For each order, we can arrange 3 people in the space with 3 seats as or , and then arrange the space with two seats as or , and then arrange the last people in the space with 1 seat, making a total of .

For case of , there are six possible orders, , , , , , . For each order, we can arrange 3 people in the two spaces with 3 seats as or or or , and then arrange the space with one seats for the people left, so there six arrangements in total: , , , , , , making a total of .

For case of , there is only one order, and each space can be arranged as or , making a total of 16.

Considering the three possibilities for seat 1, the final result is