document.write( "Question 1117698: A committee is formed from 3 seniors, 2 juniors, 2 sophomores, and 1 freshmen. All students are seated around a circular table.\r
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Algebra.Com's Answer #732867 by ikleyn(52792) $\"\"$ $\"About$

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document.write( "Assume that the seats are all numbered sequentially from 1 to 8 and placed around the circular table in order in clockwise direction.\r\n" );
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document.write( "Next assume that the freshmen occupies the chair #1.\r\n" );
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document.write( "Then you may think, as a first approximation, that each of the three groups (3 seniors / 2 juniors / 2 sophomores) represents one object.\r\n" );
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document.write( "Then you have 3 objects, and there are 3! = 1*2*3 = 6 permutations to order them.       (1)\r\n" );
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document.write( "Inside of each group you have 3! = 6 ways to order seniors,                             (2)\r\n" );
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document.write( "                              2! = 2 ways to order juniors  and                         (3)\r\n" );
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document.write( "                              2! = 2 ways to order sophomores,                          (4)\r\n" );
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document.write( "and these interior orderings are independent.\r\n" );
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document.write( "Hence, the final answer is  \r\n" );
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document.write( "(6 permutations of (1)) * (6 permutations of (2) ) * (2 permutations of (3) ) *(2 permutations of (4) ) = 6*6*2*2 = 144.\r\n" );
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document.write( "Answer.  There are 144 ways (144 circular permutations) to do it.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Thanks to this single freshmen, who facilitated the solution so much !\r
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\n" ); document.write( "\n" ); document.write( "On circular permutations, see the lesson\r
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