In how many ways can 5 boys and 3 girls be seated on 8

Consider the following arrangement:

*B*B*B*B*B*

Now, if girls occupy the places of 6 stars no girls will be together.

# of ways 3 girls can occupy the places of these 6 stars is \(C^3_6=20\);
# of ways 3 girls can be arranged on these places is \(3!=6\);
# of ways 5 boys can be arranged is \(5!=120\).

So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is \(20*6*120=14,400\).

Answer: B.
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Bunuel,

Here's what I thought:

_ O _ O _ O _ O

3 G can occupy any of the 4 "_" positions in 4C3 ways.

Similar Girls could also occupy any of the 4 "O" positions in 4C3 ways.

Boys can be permuted in 5! ways. Girls - 3! Therefore arrangements = 4C3* 2 * 5! * 3! = 4*2*120*6= 5760. Can you please let me know what I am missing?

Thanks

Please ignore. I got it. 4C3 should be 6C3.

Arrangemnets = 6C3 * 5! * 3! = B...thanks

Small typo: \(20*6*120=14,000\) should be 14,400.
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Let us take opposite of the constraint.

2 girls sitting together: :

1 case is GGBGBBBB.
Total number of ways=3!*5!*5 with just shifting the rightmost girl.
Then the 2 leftmost girls can shift one position , and using the above reasoning, the total number of ways = 3!*5!*4 and so on till the rightmost girl has 1 position.

So total number of ways = 3!*5!(5+4+3+2+1)=120*90=10800

Similarly another case is:
GBGGBBBB.
Using the above reasoning, the total number of cases is: 3!*5!*(15) =10800

Let us take 3 girls sitting together

GGGBBBBB
There are 3! *5! Ways. The 3 leftmost girls can shift 6 positions. So there are a total of 3!*5!*6=4320 ways

So total is 2*10800 + 4320=25920

The total number of possibilities = 8! Ways =40,320
Answer is 40320-25920=14400
Hence B.
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[quote="Bunuel"][quote="chetan2u"][quote="Bunuel"][quote="voodoochild"]In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?


A 5760
B 14400
C 480
D 56
E 40320



Hi Bunuel,

What is the error in this solutions : 8! - 6!3!

Total ways - girls together = not together

Regards,

Hi,
the Q asks to negate even when two are together..
What you have used is only when all three are together..
that is why your answer, 36000, is more than the actual, 14400..

Hope you have realized where you have gone wrong..
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hello chetan2u@

I want to know how can it be done the other way round,subtracting from the total number of cases:
Say we have 3 girls:G1 G2 and G3.
In the first case,we can take G1 and G2 as a single element,so it will be B1B2B3B4B5G3(G1G2)=7!*2!
Similarly we can take for G2 and G3,considering them as a single element:B1B2B3B4B5G1(G2G3)=7!*2!
And for G1 and G3,considering them as a single element,we again have 7!*2!
Total number of cases=8!
So,Cases where no two girls are together=8!-7!*2!*3=10080.

I am unable to understand,what is wrong in this approach?

I am not sure in this kind of problems why don't we consider the following arrangements : BGBBGBGB, BBGBGBGB etc when two of the boys are together

Notice that we have already included these scenarios out of 14,400.

See, Bunuel has already included 5 places for Boys and 6 places for Girls, while we have 8 people in total.
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Thanks. So the logic is whenever the space that was left out for girls to occupy is vacant then the two adjacent boys are actually together. Was kind of difficult to see unless I formulated it in my own words. That is the trick with P&C and probability, kind of seems obvious if you have nailed it, but requires a bit of imagination in some easy ones too.

# of ways 3 girls can occupy the places of these 6 stars is [m]C^3_6=20

How did you get this step??


Sent from my iPhone using GMAT Club Forum mobile app

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We have 6 places for the girls to sit and we have only 3 girls. So, Girls could select the 3 places as [m]C^3_6=20\);
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VeritasKarishma AjiteshArun egmat Bunuel
chetan2u

Why cannot we use the below approach ??

Total - when 2 girls sit together

Total--> 8!


When Two girls sit together-->> 5B 1G (G1G2)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

What is wrong with this approach.

I am not sure what you have done here - there are 3 girls and no two should sit together.
This means that G1 and G2 should not be together, G2 and G3 should not be together and G1 and G3 should not be together.

So you make the boys sit in 5 chairs. Now the girls can put their chairs in 6 spots -
S B1 S B2 S B3 S B4 S B5 S

You arrange the boys in 5! ways. Select 3 spots for girls in 6C3 ways and arrange the girls in 3! ways.

Total = 5! * 6C3 * 3! = 120 * 20 * 6 = 14400

VeritasKarishma chetan2u generis AjiteshArun Bunuel

Cannot we solve it like this??

Approach :- (Total cases - Cases when 2 girls sit together)

Total Cases :- 8!

When Two girls sit together-->> 5B G(GG)--> 7!*3C2*2!

Answer = (8!)-( 7!*3C2*2!)

However, I get different answer.

Can you please explain where am I going wrong in this case.

You will get your answer lesser than actual because there is repetition in what you are subtracting.
Say girls are a,b,c.. so you choose B,B,(a,b),c,B,B...
When you choose b,c, one combination will again be B,B,a,(b,c),B,B...
So both are same but are subtracting it twice
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[quote="voodoochild"]In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together?

A. 5760
B. 14400
C. 480
D. 56
E. 40320

[spoiler=]HEre's what I did :

girls can be made to sit
_b_b_b_b_b_
total 6 places to choose for girls ; 6c3 in 3! ways
and boys 5! ways
5!*6c3*3! = 14400
IMO B

As chetan2u mentioned above, you are double counting the eliminations. Whenever you need to arrange people such that no 2 sit together, one needs to arrange the others and place these people in between the others.