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Counting

by jrachetti » Mon Jun 15, 2009 11:28 pm
Someone could help me with this COUNTING question?

In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

A. 240
B. 360
C. 480
D. 600
E. 720


I know that 720 is the total number of posibilities without restrictions but including the condition described above?

Thanks
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by sanju09 » Tue Jun 16, 2009 12:57 am
You are correct that in the absense of restrictions, there are 6! = 720 ways in total. A quicker method is to find the various number of ways where Ann and Bea are sitting next to each other, and then take it out of 720 to get the number of ways in which Ann and Bea cannot be seated next to each other. For this let's take Ann and Bea as a single unit which can be permuted within them in 2! ways. This leaves us with five distinct units to be permuted now, total number of ways

= 720 - (5!) (2!)

= 480.

Go with C
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