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arrangement

by gupta.amit3 » Sun Nov 09, 2008 8:41 am
94. Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

1) in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats

Pls help with explanation.
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by sgarnepudi » Sun Nov 09, 2008 9:15 am
is the ans is 14?
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by stop@800 » Sun Nov 09, 2008 10:28 am
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

Total ways
2 * 4! * 4!

first is boy
patterm
BGBGBGBG
so boys can arrange in 4! ways and girls in 4! ways

first can be girl also hence again 4! * 4! ways

total
2 * 4! * 4!




in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats

lets take the case whn J and S are together

JS BGBGBG
so 3 boys and 3 girls left
3! * 3! arrangements

JS can be SJ also hence
3! * 3! * 2

now JS has 4 similar positions
so
3! * 3! * 2 * 4

now cases whn they are not together
2 * 4! * 4! - 3! * 3! * 2 * 4
=16 * 6 * 6
=576


OA please
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by 4meonly » Mon Nov 10, 2008 10:08 am
I think it will be 2*7*3!*3! with answer 502
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by mals24 » Mon Nov 10, 2008 10:50 am
@4meonly

Your reasoning is partially right. You haven't completed the solution.

In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.

Total arrangements possible

BGBGBGBG

4*4*3*3*2*2*1*1 = 4!*4! = 576

You can also have GBGBGBGB = 4!*4! = 576

Hence total arrangements = 576 + 576 = 1152

Number of ways John and Susan can sit together

Let JS be one entity

Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36

JS BGBGBG can be arranges in 7 ways = 36*7= 252

We can also have JS GBGBGB = 7*3!*3! = 252

Total arrangements = 252+252 = 504

2*7*3!*3! = 504 is the number of ways John and Susan can sit together.

We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.
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by logitech » Mon Nov 10, 2008 11:29 am
mals24 wrote:@4meonly

Your reasoning is partially right. You haven't completed the solution.

In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.

Total arrangements possible

BGBGBGBG

4*4*3*3*2*2*1*1 = 4!*4! = 576

You can also have GBGBGBGB = 4!*4! = 576

Hence total arrangements = 576 + 576 = 1152

Number of ways John and Susan can sit together

Let JS be one entity

Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36

JS BGBGBG can be arranges in 7 ways = 36*7= 252

We can also have JS GBGBGB = 7*3!*3! = 252

Total arrangements = 252+252 = 504

2*7*3!*3! = 504 is the number of ways John and Susan can sit together.

We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.
Mals,

Good work. What most of the people did not see in this problem was John and Susan can sit in 7 different positions, and since they can switch = 14 ways

JS XX XX XX
XJ SX XX XX
XX JS XX XX
XX XJ SX XX
XX XX JS XX
XX XX XJ SX
XX XX XX JS

so it comes down to:

2x4!x4! - 2x7x3!x3! = 648
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by cramya » Mon Nov 10, 2008 2:35 pm
Nice solution Mals!

Your reasoning is partially right. You haven't completed the solution.

In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.

Total arrangements possible

BGBGBGBG

4*4*3*3*2*2*1*1 = 4!*4! = 576

You can also have GBGBGBGB = 4!*4! = 576

Hence total arrangements = 576 + 576 = 1152

Number of ways John and Susan can sit together

Let JS be one entity

Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36

JS BGBGBG can be arranges in 7 ways = 36*7= 252

We can also have JS GBGBGB = 7*3!*3! = 252

Total arrangements = 252+252 = 504

2*7*3!*3! = 504 is the number of ways John and Susan can sit together.

We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.





LOGITECH HOPE THIS HELPS :-)
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by stop@800 » Mon Nov 10, 2008 8:07 pm
logitech wrote:
mals24 wrote:@4meonly

Your reasoning is partially right. You haven't completed the solution.

In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.

Total arrangements possible

BGBGBGBG

4*4*3*3*2*2*1*1 = 4!*4! = 576

You can also have GBGBGBGB = 4!*4! = 576

Hence total arrangements = 576 + 576 = 1152

Number of ways John and Susan can sit together

Let JS be one entity

Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36

JS BGBGBG can be arranges in 7 ways = 36*7= 252

We can also have JS GBGBGB = 7*3!*3! = 252

Total arrangements = 252+252 = 504

2*7*3!*3! = 504 is the number of ways John and Susan can sit together.

We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.
Mals,

Good work. What most of the people did not see in this problem was John and Susan can sit in 7 different positions, and since they can switch = 14 ways

JS XX XX XX
XJ SX XX XX
XX JS XX XX
XX XJ SX XX
XX XX JS XX
XX XX XJ SX
XX XX XX JS
But first condition ("4 boys and 4 girls can be seated alternatively") will get violated in some cases.

so it comes down to:

2x4!x4! - 2x7x3!x3! = 648
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by logitech » Mon Nov 10, 2008 8:18 pm
stop@800 wrote:
logitech wrote:
mals24 wrote:@4meonly

Your reasoning is partially right. You haven't completed the solution.

In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.

Total arrangements possible

BGBGBGBG

4*4*3*3*2*2*1*1 = 4!*4! = 576

You can also have GBGBGBGB = 4!*4! = 576

Hence total arrangements = 576 + 576 = 1152

Number of ways John and Susan can sit together

Let JS be one entity

Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36

JS BGBGBG can be arranges in 7 ways = 36*7= 252

We can also have JS GBGBGB = 7*3!*3! = 252

Total arrangements = 252+252 = 504

2*7*3!*3! = 504 is the number of ways John and Susan can sit together.

We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.
Mals,

Good work. What most of the people did not see in this problem was John and Susan can sit in 7 different positions, and since they can switch = 14 ways

JS XX XX XX
XJ SX XX XX
XX JS XX XX
XX XJ SX XX
XX XX JS XX
XX XX XJ SX
XX XX XX JS
But first condition ("4 boys and 4 girls can be seated alternatively") will get violated in some cases.

so it comes down to:

2x4!x4! - 2x7x3!x3! = 648
That's good point. And what do you offer as a solution to your question ?
LGTCH
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by cramya » Mon Nov 10, 2008 8:20 pm
I am waiting for the business man solution /approach, Logitech and who better to provide than yourself :-)

Dont have any Bud?
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by logitech » Mon Nov 10, 2008 8:23 pm
cramya wrote:I am waiting for the business man solution /approach, Logitech and who better to provide than yourself :-)

Dont have any Bud?
Ladies first
LGTCH
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by cramya » Mon Nov 10, 2008 8:24 pm
Go then. What r u waiting for? We will follow you....
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by logitech » Mon Nov 10, 2008 8:33 pm
stop@800 wrote:


But first condition ("4 boys and 4 girls can be seated alternatively") will get violated in some cases.
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

Okay, I re-read the stem again. You have a great point there regarding the violation but actually the stem itself limits its scope. We are only dealing with alternative seatings. So we are good!
LGTCH
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