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# Eight women of eight different heights are to pose for a

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CEO
Joined: 15 Aug 2003
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Eight women of eight different heights are to pose for a [#permalink]

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07 Jan 2004, 01:07
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Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many diffrent ways can the women pose?

(A) 2
(B) 14
(C) 15
(D) 16
(E) 18

Director
Joined: 13 Nov 2003
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Location: BULGARIA
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08 Jan 2004, 04:57
In my opinion the answer is A) 2, 1,2,3,4,5,6,7,8 are the women ordered in height from left to right. Fist row 1,2,3,4 second row 5,6,7,8 one way as 5 is behind 1 and so on. Second way first row 1,3,5,7, second row 2,4,6,8, as 2 is behind 1 and so on . The women are ordered in increasing height from left to right and the taller are behind the shorter
SVP
Joined: 30 Oct 2003
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09 Jan 2004, 20:44
let the height of each peron be represented as
x+0, x+1, x+2, x+3, x+4, x+5, x+6, x+7

If first row is
x+0,x+1,x+2,x+3,x+4 then 4 people behind them can stand in 4! ways

if first row second row is arranged as

x+0, x+2, x+4, x+6
x+1, x+3,x+5, x+7

here x+1 and x+2 can be exchanged x+3 and x+4 can ge exchanged x+5 and x+6 can be exchanged
These are the only possible ways

so we have 4! + 1 + 3 = 16
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