document.write( "Question 1206384: There are 6 people who will sit in a row but out of them, Ronnie, will always be left of Annie and Rachel will always be right of Annie. How many row arrangements are possible?\r
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Algebra.Com's Answer #843825 by math_tutor2020(3817) ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: 120\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Explanation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A,B,C = Ronnie, Annie, Rachel \n" ); document.write( "D,E,F = the other 3 people\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A must be to the left of B. \n" ); document.write( "C must be to the right of B. \n" ); document.write( "In other words: we must have A__B__C where 0 or more letters will go in the blanks.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Imagine that persons D,E, and F will select numbers 1 through 6 from a hat. The selections are done without replacement. The numbers refer to seating arrangement. \n" ); document.write( "1 = left most seat \n" ); document.write( "6 = right most seat \n" ); document.write( "Use the nCr combination formula (or Pascals Triangle) to determine there are p = 6C3 = 20 different ways to do this where order doesn't matter. \n" ); document.write( "Eg: Group {1,3,4} is the same as {3,1,4}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Once persons D,E, and F find their seat, there are q = 3! = 3*2*1 = 6 ways to permute the 3 members of that group. \n" ); document.write( "Think of it like a game of musical chairs except none of the chairs get taken away and everyone is able to get a seat.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, we have p*q = 20*6 = 120 ways to ensure that the ordering is A,B,C where there may or may not be a gap between Ronnie, Annie, Rachel.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A few examples of valid arrangements: \n" ); document.write( "ABC,DEF \n" ); document.write( "ABD,CEF \n" ); document.write( "ADB,ECF\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Examples that aren't valid arrangements \n" ); document.write( "CAB,DEF \n" ); document.write( "CAD,FEB \n" ); document.write( "Both aren't valid because C is not to the right of B. \n" ); document.write( " \n" ); document.write( " |