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Question

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.

sgarnepudi · Answer

is the ans is 14?

stop@800 · Answer

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

4meonly · Answer

I think it will be 2*7*3!*3! with answer 502

mals24 · Answer

@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.

logitech · Answer

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
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

cramya · Answer

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

stop@800 · Answer

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

logitech · Answer

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 ?

cramya · Answer

I am waiting for the business man solution /approach, Logitech and who better to provide than yourself Dont have any Bud?

logitech · Answer

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

cramya · Answer

Go then. What r u waiting for? We will follow you....

logitech · Answer

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!

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