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Determine the number of integer solutions of x1 + x2 + x3 = 6 where
a. xi > 0, 1 <= i <= 3
b. x1x2 >= 2 and x3 > 0
c. xi > 2, 1 <= i <= 3
the numbers and leter I are to be substrips.
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So...
Last edited by simron (2008-05-12 06:18:09)
Linux FTW
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part a.) (m-1)(m-2) / 2 ways, where m = 6, so (5)(4)/2 = 10 ways.
(This is just like my pie sharing post in this is cool with 3 people!!)
part b.) e.g. (-2)+(-2)+(10) e.g. (-5)+(-7)+(18)
infinite solutions if product of x1 and x2 go positive from 2 negatives.
part c.)
x1 x2 x3
no solutions because 3 + 3 + 3 > 6
igloo myrtilles fourmis
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