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Combinatorics with restrictions

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Source: — Problem Solving |

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by Brent@GMATPrepNow » Tue Dec 31, 2013 10:02 am
dddanny2006 wrote:Steve goes to the movies with four friends but refuses to sit next to one of
them. In how many ways can they be arranged?

Is the answer 5!-4!2! ??

Thanks
Exactly.

We'll use the fact that # acceptable arrangements = # total arrangements (ignoring restriction) - # bad arrangements

# total arrangements (ignoring restriction)
If we ignore the restriction, we are simply arranging the 5 friends.
This can be accomplished in 5! different ways (since we can arrange n unique objects in n! ways)

# bad arrangements
To count the arrangements where Steve and his "enemy" are seated together, let's GLUE them together to create ONE single entity that we'll call "Steve+Enemy" (where Steve is on the left and his enemy is on the right)
In how many ways can we arrange "Steve+Enemy" and the 3 other friends?
Well, we have 4 unique objects, so we can arrange them in 4! ways.

IMPORTANT: Notice that we can also glue the two men to get "Enemy+Steve" (where the Enemy is on the left and Steve is on the right)
We can arrange "Enemy+Steve" and the 3 other friends, in 4! ways.


So, # acceptable arrangements = # total arrangements (ignoring restriction) - # bad arrangements
= 5! - (4! + 4!)
= 5! - 2(4!)
= 120 - 2(24)
= 72

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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