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:

$8 = 1 + 1 + 1 + 5 = 1 + 1 + 2 + 4 = 1 + 2 +2 + 3 = 1 + 1 + 3 + 3 = 2 + 2 + 2 + 2$ .

For case of $1+1+1+5$ , there are four possible orders. For each order, we can arrange 5 people in the space with 5 seats as $BCBCB$ or $CBCBC$ , and then arrange two B's or two C's in two of three remaining spaces, making a total of $2\times 3\times = 24$ .

For case of $1+1+2+4$ , there are twelve possible orders, $1124$ , $1241$ , $2411$ , $4112$ , $1142$ , $1421$ , $4211$ , $2114$ , $1214$ , $2141$ , $1412$ , $4121$ . For each order, we can arrange 4 people in the space with 4 seats as $BCBC$ or $CBCB$ , and then arrange the space with two seats as $BC$ or $CB$ , and then choose 1 space for $B$ and one space for $C$ , making a total of $2\times 2\times 2\times 12 = 96$ .

For case of $1+2+2+3$ , 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 $BCB$ or $CBC$ , and then arrange the space with two seats as $BC$ or $CB$ , and then arrange the last people in the space with 1 seat, making a total of $2\times 2\times 2\times 12= 96$ .

For case of $1+1+3+3$ , there are six possible orders, $1133$ , $1331$ , $3311$ , $3113$ , $1313$ , $3131$ . For each order, we can arrange 3 people in the two spaces with 3 seats as $BCB/BCB$ or $BCB/CBC$ or $CBC/BCB$ or $CBC/CBC$ , and then arrange the space with one seats for the people left, so there six arrangements in total: $BCB/BCB/C/C$ , $BCB/CBC/B/C$ , $BCB/CBC/C/B$ , $CBC/BCB/B/C$ , $CBC/BCB/C/B$ , $CBC/CBC/B/B$ , making a total of $6\times 6 = 36$ .