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Combinations Problem

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Manager
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Joined: 22 Nov 2003
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Location: New Orleans
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Combinations Problem [#permalink] New post 04 Dec 2003, 11:18
I posted this question in a reply, but I think this question is better served by being offered as a new thread. Thanks for understanding.
---

If we have 7 people in a movie theater, and 3 people choose not to sit next to each other, how many arrangements do we have?

7! = total possibilities = 5040

To fill all 7 seats, we take 5!, b/c in this example 1 unit is now 3 (giving us 7 filled seats)
5! = 120

120 * 3! b/c there are 6 different possibilities for 3 people to be seated next to each other.

I.e. 3 people represented by a,b,c

abc
acb
bac
bca
cab
cba

120 * 6 = 720 arrangements where these 3 jokers sit next to each other

... so ...

5040 - 720 = 4320 total possibilities

Is this correct?

Thanks,
CJ
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Re: Combinations Problem [#permalink] New post 04 Dec 2003, 12:27
csperber wrote:

If we have 7 people in a movie theater, and 3 people choose not to sit next to each other, how many arrangements do we have?




Total ways of seating 7 people in a row = 7 !

if the three sit together,

consider these three as a single unit , we will now have 5 people...these five people can sit in 5! ways...also the three people that are together can themselves sit in 3! ways...

thats a total of 5! * 3!

total ways = 7! - 5! * 3!
SVP
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 [#permalink] New post 05 Dec 2003, 05:00
agree
7!–[5!*3!]
  [#permalink] 05 Dec 2003, 05:00
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