document.write( "Question 943719: Suppose that 10 identical whiteboards are to be divided among 3 schools, SB, SPIA and IT, how many divisions are possible? How many if SB must receive at least 2 whiteboards. \n" ); document.write( "
Algebra.Com's Answer #575545 by Edwin McCravy(20055)\"\" \"About 
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document.write( "The formula is (n+r-1)C(r-1) with n=10-3*2 = 4 whiteboards left after giving\r\n" );
document.write( "each school 2, r = 3, the number of schools.  (4+3-1)C(3-1) = 6C2 = 15.\r\n" );
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document.write( "Here's why that works:\r\n" );
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document.write( "Let the schools br A, B, and C.\r\n" );
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document.write( "First give each of the 3 schools 2 whiteboards each, which guarantees\r\n" );
document.write( "that each school will get at least 2 whiteboards.\r\n" );
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document.write( "Then there are 4 whiteboards left to distribute to one, two or all 3 \r\n" );
document.write( "of the schools in addition to the 2 they already have. \r\n" );
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document.write( "To figure this out, get (or imagine) 4+3= 7 slips of paper.  Lay them\r\n" );
document.write( "left to right.  Write \"School C\" on the slip of paper on the far\r\n" );
document.write( "right.\r\n" );
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document.write( "Now choose any 2 of the other 6 sheets of paper.  Write School A on the\r\n" );
document.write( "slip of paper you just chose which is farther to the left, and write \r\n" );
document.write( "\"School B\" on the other one you chose, which is farther to the right.\r\n" );
document.write( "You can do this in 6C2 = 15 ways.  This is the answer to the problem.\r\n" );
document.write( "Why: OK, here is why:\r\n" );
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document.write( "You have 7 slips of paper laid out left to right.  The rightmost one has \r\n" );
document.write( "\"School C\" written on it.  One of the other slips of paper have \"School A\"\r\n" );
document.write( "written on it. And one between it and the right end has \"School B\" written on\r\n" );
document.write( "it.  The remaining 4 slips of paper have nothing written on them.  The number\r\n" );
document.write( "of slips of paper to the left of the slip that reads \"School A\"  is the\r\n" );
document.write( "number of whiteboards to give to school A in addition to the 2 it already\r\n" );
document.write( "has.  (If the slip that reads School A\" is on the left end, then there are no\r\n" );
document.write( "slips of paper left of it.  Then the number of slips of paper left of it is 0,\r\n" );
document.write( "so in that case School A gets 0 or NO additional whiteboards.\r\n" );
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document.write( "The number of slips of paper between the slips that reads \"School A\" and\r\n" );
document.write( "\"School B\" is the number of whiteboards to give to school B in addition to the 2\r\n" );
document.write( "it already has.  (If the slip that reads School B\" happens to be adjacent to the\r\n" );
document.write( "slip that reads \"School A\", then there are no slips of paper between them.  Then\r\n" );
document.write( "the number of slips of paper between them is 0, so in that case School B gets 0\r\n" );
document.write( "or NO additional whiteboards.\r\n" );
document.write( "  \r\n" );
document.write( "The number of slips of paper between the slips that reads \"School B\" and\r\n" );
document.write( "\"School C\" on the right end is the number of whiteboards to give to school C in\r\n" );
document.write( "addition to the 2 it already has.  (If the slip that reads School B\" happens to\r\n" );
document.write( "be adjacent to the slip that reads \"School C\", then there are no slips of paper\r\n" );
document.write( "between them.  Then the number of slips of paper between them is 0, so in that\r\n" );
document.write( "case School C gets 0 or NO additional whiteboards.\r\n" );
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document.write( "Each choice of those 2 slips of paper determines a different partition of of\r\n" );
document.write( "the remaining 4 whiteboards.\r\n" );
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document.write( "Answer:  (n+r-1)C(r-1), n=10-3*2 = 4 (4+3-1)C(3-1) = 6C2 = 15.\r\n" );
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document.write( "Edwin
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