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3 girls and 3 boys sit down in in the row of 7 chairs. If

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Senior Manager
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3 girls and 3 boys sit down in in the row of 7 chairs. If [#permalink] New post 20 Jan 2006, 15:17
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3 girls and 3 boys sit down in in the row of 7 chairs. If no girl will sit next to another girl and no boy will sit next to another boy, in how many different ways can the girls and boys sit.
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 [#permalink] New post 20 Jan 2006, 16:46
Is it 7 * 3! *3!

We have to take empty seat also as an arrangement.
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 [#permalink] New post 20 Jan 2006, 21:49
(g b g b g b _) or (b g b g b g _)
(g b g b g _ b) or (b g b g b _ g)

etc

So total = 3! * 3! * 7 = 252
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 [#permalink] New post 21 Jan 2006, 09:38
I am slightly confused by the approach given above. Let us assume there is 1 girl and 1 boy and 3 seats.

This provides 6 arrangements

_gb
_bg
b_g
g_b
bg_
gb_

For this the answer should be 1!*P(3,2)=6. Similarly according to me the answer for original question should be 3!*P(4,3) = 144
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 [#permalink] New post 21 Jan 2006, 10:55
I think it is 2*7*3!*3!

If we take the arrangement: bgbgbg_
there are 7 choices for "_"

Also if we take gbgbgb_, there are 7 choices for the "_"

And for each of the two cases above, there are 3!*3! arrangements for boys and girs

Total choices = 2*7*3!*3!

OA please?
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 [#permalink] New post 25 Jan 2006, 04:26
I am pathetic at combos, can someone tell me why we are multiplying by 7.
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 [#permalink] New post 25 Jan 2006, 10:16
shahnandan wrote:
I am pathetic at combos, can someone tell me why we are multiplying by 7.


because of 1 empty chair that can be at 7 different places
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  [#permalink] 25 Jan 2006, 10:16
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3 girls and 3 boys sit down in in the row of 7 chairs. If

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