document.write( "Question 166109: How many different seating arrangements are possible for 6 people in 4 chairs? \n" ); document.write( "
Algebra.Com's Answer #122392 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "Since order does not matter (ie the seats have no unique positions), we must use the combination formula:\r
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\n" ); document.write( "\n" ); document.write( "\"n%21%2F%28n-r%29%21r%21\" Start with the combination formula.\r
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\n" ); document.write( "\n" ); document.write( "\"6%21%2F%286-4%29%214%21\" Plug in \"n=6\" and \"r=4\"\r
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\n" ); document.write( "\n" ); document.write( "\"6%21%2F2%214%21\" Subtract \"6-4\" to get 2\r
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\n" ); document.write( "\n" ); document.write( "Expand 6!
\n" ); document.write( "\"%286%2A5%2A4%2A3%2A2%2A1%29%2F2%214%21\"\r
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\n" ); document.write( "\n" ); document.write( "Expand 2!
\n" ); document.write( "\"%286%2A5%2A4%2A3%2A2%2A1%29%2F%282%2A1%294%21\"\r
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\n" ); document.write( "\n" ); document.write( "\"%286%2A5%2A4%2A3%2Across%282%2A1%29%29%2F%28cross%282%2A1%29%294%21\" Cancel\r
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\n" ); document.write( "\n" ); document.write( "\"%286%2A5%2A4%2A3%29%2F4%21\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "Expand 4!
\n" ); document.write( "\"%286%2A5%2A4%2A3%29%2F%284%2A3%2A2%2A1%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"360%2F%284%2A3%2A2%2A1%29\" Multiply 6*5*4*3 to get 360\r
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\n" ); document.write( "\n" ); document.write( "\"360%2F24\" Multiply 4*3*2*1 to get 24\r
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\n" ); document.write( "\n" ); document.write( "\"15\" Now divide\r
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\n" ); document.write( "\n" ); document.write( "So 6 choose 4 (where order doesn't matter) yields 15 unique combinations. So there are 15 possible ways that 6 people can sit in 4 chairs.
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