https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_24&feed=atom&action=history 2019 Mock AMC 10B Problems/Problem 24 - Revision history 2021-04-15T03:26:02Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_24&diff=142901&oldid=prev Mathsuper: Solution 1: 2021-01-21T06:57:05Z <p>Solution 1:</p> <p><b>New page</b></p><div>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:<br /> <br /> &lt;math&gt;8 = 1 + 1 + 1 + 5 = 1 + 1 + 2 + 4 = 1 + 2 +2 + 3 = 1 + 1 + 3 + 3 = 2 + 2 + 2 + 2&lt;/math&gt;.<br /> <br /> For case of &lt;math&gt;1+1+1+5&lt;/math&gt;, there are four possible orders. For each order, we can arrange 5 people in the space with 5 seats as &lt;math&gt;BCBCB&lt;/math&gt; or &lt;math&gt;CBCBC&lt;/math&gt;, and then arrange two B's or two C's in two of three remaining spaces, making a total of &lt;math&gt;2\times 3\times = 24&lt;/math&gt;.<br /> <br /> For case of &lt;math&gt;1+1+2+4&lt;/math&gt;, there are twelve possible orders, &lt;math&gt;1124&lt;/math&gt;, &lt;math&gt;1241&lt;/math&gt;, &lt;math&gt;2411&lt;/math&gt;, &lt;math&gt;4112&lt;/math&gt;, &lt;math&gt;1142&lt;/math&gt;, &lt;math&gt;1421&lt;/math&gt;, &lt;math&gt;4211&lt;/math&gt;, &lt;math&gt;2114&lt;/math&gt;, &lt;math&gt;1214&lt;/math&gt;, &lt;math&gt;2141&lt;/math&gt;, &lt;math&gt;1412&lt;/math&gt;, &lt;math&gt;4121&lt;/math&gt;. For each order, we can arrange 4 people in the space with 4 seats as &lt;math&gt;BCBC&lt;/math&gt; or &lt;math&gt;CBCB&lt;/math&gt;, and then arrange the space with two seats as &lt;math&gt;BC&lt;/math&gt; or &lt;math&gt;CB&lt;/math&gt;, and then choose 1 space for &lt;math&gt;B&lt;/math&gt; and one space for &lt;math&gt;C&lt;/math&gt;, making a total of &lt;math&gt;2\times 2\times 2\times 12 = 96&lt;/math&gt;.<br /> <br /> For case of &lt;math&gt;1+2+2+3&lt;/math&gt;, 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 &lt;math&gt;BCB&lt;/math&gt; or &lt;math&gt;CBC&lt;/math&gt;, and then arrange the space with two seats as &lt;math&gt;BC&lt;/math&gt; or &lt;math&gt;CB&lt;/math&gt;, and then arrange the last people in the space with 1 seat, making a total of &lt;math&gt;2\times 2\times 2\times 12= 96&lt;/math&gt;.<br /> <br /> For case of &lt;math&gt;1+1+3+3&lt;/math&gt;, there are six possible orders, &lt;math&gt;1133&lt;/math&gt;, &lt;math&gt;1331&lt;/math&gt;, &lt;math&gt;3311&lt;/math&gt;, &lt;math&gt;3113&lt;/math&gt;, &lt;math&gt;1313&lt;/math&gt;, &lt;math&gt;3131&lt;/math&gt;. For each order, we can arrange 3 people in the two spaces with 3 seats as &lt;math&gt;BCB/BCB&lt;/math&gt; or &lt;math&gt;BCB/CBC&lt;/math&gt; or &lt;math&gt;CBC/BCB&lt;/math&gt; or &lt;math&gt;CBC/CBC&lt;/math&gt;, and then arrange the space with one seats for the people left, so there six arrangements in total: &lt;math&gt;BCB/BCB/C/C&lt;/math&gt;, &lt;math&gt;BCB/CBC/B/C&lt;/math&gt;, &lt;math&gt;BCB/CBC/C/B&lt;/math&gt;, &lt;math&gt;CBC/BCB/B/C&lt;/math&gt;, &lt;math&gt;CBC/BCB/C/B&lt;/math&gt;, &lt;math&gt;CBC/CBC/B/B&lt;/math&gt;, making a total of &lt;math&gt;6\times 6 = 36&lt;/math&gt;.<br /> <br /> For case of &lt;math&gt;2+2+2+2&lt;/math&gt;, there is only one order, and each space can be arranged as &lt;math&gt;BC&lt;/math&gt; or &lt;math&gt;CB&lt;/math&gt;, making a total of 16.<br /> <br /> Considering the three possibilities for seat 1, the final result is<br /> &lt;cmath&gt;<br /> 3\times(24+96+96+36+16) = \boxed{804}.<br /> &lt;/cmath&gt;</div> Mathsuper