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by Options » Sun Jul 03, 2011 8:35 am
sigh i'm stuck with this question some1 help!

Jane, Bryan, Sheryl, David and Ben went to a concert. Ben has a crush on Sheryl, while Bryan is not on speaking terms with David. How many arrangements are possible if
1) Ben insists on sitting next to Sheryl
2) Bryan refuses to sit next to David
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by Frankenstein » Sun Jul 03, 2011 8:47 am
Hi,
Let Ben and Sheryl be grouped together. So, we have 3 persons +1 group of Ben and Sheryl.
So, they can be arranged in 4! ways
group of Ben and Sheryl can be permuted in 2! ways
So, number of ways in 4!*2! = 48 ways
Now, these include Bryan and David together as well.
Let's consider the case of Bryan and David together as well. So we make them a group.
Number of ways is 3!*2!*2! = 24ways
So, for them not to be together, the number of ways is 48-24 = 24ways.

(OR)

Jane, Bryan, Sheryl, David and Ben went to a concert. Ben has a crush on Sheryl,
Consider __ Jane__(Ben+Sheryl)__
Jane and (Ben+Sheryl can be arranged in 2!*2! = 4 ways.
There are 3 gaps in which David and Bryan can be filled.
This can be done in 3P2 = 6 ways.
So, total number of ways is 4*6 = 24 ways
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by amit2k9 » Sun Jul 03, 2011 11:18 pm
questions like there you should use symmetry.

Ben and Sheryl sit together.

total people 4 (Be-Sh one group)
also Be-SH can sit in 2! ways

thus total ways = 2!*4! = 48.
in half if these 48 cases D and Br will sit apart and in half they will sit together.

thus total cases = 48/2 = 24.
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by sunilrawat » Mon Jul 04, 2011 1:12 am
amit2k9 wrote: in half if these 48 cases D and Br will sit apart and in half they will sit together.

thus total cases = 48/2 = 24.
why half the cases? please explain
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