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by zooki » Fri Oct 14, 2011 10:24 am
A, B, C, D, E,and F (6 people)go to movie and sit next to each other in 6 adjacent seats.
1. If A and B will not sit next to each other, in how many different arrangement can 6 people sit?
2. If any 2 will not sit next to each other, in how many different arrangement can 6 people sit?

Can someone please answer using permutation (if applicable)?
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by CappyAA » Fri Oct 14, 2011 10:39 am
Good = Total - Bad

The Total number of ways 6 people can be arranged is 6! or 720.

The Bad is the times where A & B are next to each other. Imagine A & B are stuck together and you have 5 people. A&B, C, D, E, and F. This group can be arranged 5! or 120 ways. This will count all the times A&B are sitting next to each other with A on one side of B. But A can also sit on the other side of B (picture the first count as the times A is to the left of B). To find the total number of ways A & B can sit next to each other, just multiply by 2 to get 240.

Now you can find the number of times A & B do not sit next to each other:

Good = Total - Bad
Good = 720 - 240
Good = 480
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by fcabanski » Fri Oct 14, 2011 10:44 am
Whenever a question involves two things together or not together, treat them as a unit (as in the above solution.)
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by zooki » Fri Oct 14, 2011 11:34 am
I think the below method is quicker.

_C_D_E_F_

we can arrange CDEF in 4!= 24 ways
for the gaps we have 5p2 ways =20 ways. [we can have A or B in any of the gaps].
Total =24*20=480 ways.
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