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# How many ways 3 girls and 4 boys can be seated so that girls

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How many ways 3 girls and 4 boys can be seated so that girls [#permalink]  16 May 2005, 11:18
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How many ways 3 girls and 4 boys can be seated so that girls never sit side by side?

My approach to this question is first to find total no. of arrangements, which is 7!. Then, first, subtract no. of possibilities when all 3 girls sit together (5!*3!) and, second, no. of possibilities when 2 girls out of 3 sit together.

Can someone help me to find the last item, i.e. no. of 2 girls out of 3 sitting together?

Unfortunately, I don't know the OA, so we need to find it on our own.
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Re: PS: Permutation [#permalink]  16 May 2005, 11:38
You don't need to consider situations where three girls sit together.
How many combinations are there where two girls sit together.
Make two girls as one.
Selecting two girls out of 3, ordered gives you 3!/1! = 3!
Then order the rest in 6! ways.
You end up with 3! * 6! ways.

The answer is 7! - 3! * 6! = 6! (7 - 6) = 6!
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Re: PS: Permutation [#permalink]  16 May 2005, 14:44
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

_________________

Who says elephants can't dance?

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Re: PS: Permutation [#permalink]  16 May 2005, 16:28
kapslock wrote:
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

You have two times more outcomes then I do.

I think I see where's my flaw.
Classic
Counting number of outcomes where two girls sit together I counted arrangements of three girls together twice.
Number of outcomes where three girls sit together as noted originally - 5! * 3! = 6!
So number of outcomes where two girls sit together is in fact -
6! * 3! - 6!
And outcomes that we want 7! - 6! * 3! + 6! = 2 * 6!
Note to self, remember rules of Venn diagrams
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4 boys can be arranged in 4! ways....

there are 5 spaces for the three girls ==>> 5C3 ways

Also the girls can be arranged in 3! ways among themselves.

4!*3!*5C3
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Re: PS: Permutation [#permalink]  17 May 2005, 10:40
kapslock wrote:
I am getting an answer of 5!x4x3 = 1440.

_ B _ B _ B _ B _
4 Boys can be rearranged in their prescribed positions in 4! ways.
3 Girls can be placed in 5 spaces in 5x4x3 ways.
Therefore, total = 4! x 5 x 4 x 3 = 5! x 4 x 3 = 1440.

kapslock, I've researched this type of problem today and come across the method used exactly the same as you described. Many thanks, guys, for insightful approaches.
Re: PS: Permutation   [#permalink] 17 May 2005, 10:40
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