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In how many arrangements can a teacher seat 3 girls and 4

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Senior Manager
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In how many arrangements can a teacher seat 3 girls and 4 [#permalink]

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New post 28 Nov 2007, 12:08
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In how many arrangements can a teacher seat 3 girls and 4 boys in a row of 7 seats if the girls are to have the secon, fourth and sixth seats?

(A) 12
(B) 36
(C) 144
(D) 288
(E) 5,040

Kudos [?]: 70 [1], given: 0

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CEO
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Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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 [#permalink]

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New post 28 Nov 2007, 12:25
C.

for girls: p=3*2*1=6
for boys: q=4*3*2*1=24

together: p*q=6*24=144

Kudos [?]: 4593 [0], given: 360

Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4593 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
 [#permalink]

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New post 28 Nov 2007, 12:32
rajpdsouza, GOOD LUCK on tomorrow exam!!! :-D

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Senior Manager
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 [#permalink]

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New post 28 Nov 2007, 12:33
walker wrote:
rajpdsouza, GOOD LUCK on tomorrow exam!!! :-D


Thanks! Although my post count is low, I have been a daily lurker since I joined. Hopefully, another gmatclubber will join the 700-ranks.

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Re: Permutation Question [#permalink]

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New post 28 Nov 2007, 12:35
lumone wrote:
In how many arrangements can a teacher seat 3 girls and 4 boys in a row of 7 seats if the girls are to have the secon, fourth and sixth seats?

(A) 12
(B) 36
(C) 144
(D) 288
(E) 5,040



BGBGBGB Girls can be places in 3! ways and boys can be placed in 4! ways. so 3!*4!=144.


C

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New post 27 Aug 2008, 02:43
This is another reasonable solution:

On seat one you can put any of the 4 boys, on seat two you can put any of the 3 girls, on seat 3 you can put any of the remaining 3 boys, on seat four you can put any of the remaining 2 girls, seat five you can put any of the remaining 2 boys, and on seats six and seven you can put the one remaining girl and boy. You can write this representation in the following form:

4x3x3x2x2x1x1 = 144

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Manager
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Re: Permutation Question [#permalink]

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New post 27 Sep 2009, 22:37
In how many arrangements can a teacher seat 3 girls and 4 boys in a row of 7 seats if the girls are to have the secon, fourth and sixth seats?

(A) 12
(B) 36
(C) 144
(D) 288
(E) 5,040



Soln: 4! * 3! ways
= 144

Ans is C

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Senior Manager
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Re: Permutation Question [#permalink]

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New post 16 Feb 2010, 13:00
lumone wrote:
In how many arrangements can a teacher seat 3 girls and 4 boys in a row of 7 seats if the girls are to have the secon, fourth and sixth seats?

(A) 12
(B) 36
(C) 144
(D) 288
(E) 5,040


BGBGBGB

G can be arranged in 3! ways
B can be arranged in 4! ways

Therefore 3! x 4! = 144
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Re: Permutation Question   [#permalink] 16 Feb 2010, 13:00
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