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

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In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]  24 Sep 2012, 12:02
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Question Stats:

55% (03:30) correct 45% (02:03) wrong based on 51 sessions
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

[Reveal] Spoiler:
HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?
[Reveal] Spoiler: OA

Last edited by voodoochild on 24 Sep 2012, 12:37, edited 1 time in total.
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Re: In how many ways can 5 boys and 3 girls [#permalink]  24 Sep 2012, 12:23
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Expert's post
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voodoochild wrote:
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

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?

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 $$20*6*120=14,000$$.

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Manager
Joined: 16 Feb 2011
Posts: 192
Schools: ABCD
Followers: 1

Kudos [?]: 103 [0], given: 78

Re: In how many ways can 5 boys and 3 girls [#permalink]  24 Sep 2012, 12:37
Bunuel wrote:
voodoochild wrote:
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

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?

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 $$20*6*120=14,000$$.

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
Manager
Joined: 16 Feb 2011
Posts: 192
Schools: ABCD
Followers: 1

Kudos [?]: 103 [0], given: 78

Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]  24 Sep 2012, 12:59
Please ignore. I got it. 4C3 should be 6C3.

Arrangemnets = 6C3 * 5! * 3! = B...thanks
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
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Re: In how many ways can 5 boys and 3 girls [#permalink]  24 Sep 2012, 13:32
Bunuel wrote:
voodoochild wrote:
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

HEre's what I did :

5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct?

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 $$20*6*120=14,000$$.

Small typo: $$20*6*120=14,000$$ should be 14,400.
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Re: In how many ways can 5 boys and 3 girls be seated on 8 [#permalink]  08 Jan 2015, 00:59
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Re: In how many ways can 5 boys and 3 girls be seated on 8   [#permalink] 08 Jan 2015, 00:59
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