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# For a finite sequence of nonzero numbers, the number of

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VP
Joined: 22 Nov 2007
Posts: 1104
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Kudos [?]: 215 [0], given: 0

For a finite sequence of nonzero numbers, the number of [#permalink]  18 Dec 2007, 09:39
For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms in the sequence for which the product of the 2 consecutive term is negative. What is the number of varations in sign for the sequence 1, -3, 2, 5, -4, -6?

A. 1
B. 2
C. 3
D. 4
E. 5
CEO
Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 414

Kudos [?]: 2244 [0], given: 359

Expert's post
C

I think It is not combination.

1, -3, 2, 5, -4, -6
1, -3, 2, 5, -4, -6
1, -3, 2, 5, -4, -6
CEO
Joined: 29 Mar 2007
Posts: 2585
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Kudos [?]: 244 [0], given: 0

Re: combinations? [#permalink]  19 Dec 2007, 14:34
marcodonzelli wrote:
For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms in the sequence for which the product of the 2 consecutive term is negative. What is the number of varations in sign for the sequence 1, -3, 2, 5, -4, -6?

A. 1
B. 2
C. 3
D. 4
E. 5

Im really confused by the definition of consecutive term here. b/c i dont see any real consecutive terms here at all.

but if we were suppose to consider real consecutive terms: -1,0,1 etc.. there wouldnt be any negative variations...

so it must mean terms such as this from the list above:

-6*5
-4*3
-3*2

essentially 6 comes after 5 and we just ignore then minus sign to apply the consecutive term... i prolly dont make much sense, but I think its C
Re: combinations?   [#permalink] 19 Dec 2007, 14:34
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