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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:

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, 12:00

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