Man.. how you do this discreet math problem ...

[TurtleMan]

Man as much as i hate to post in OT about my discreet math problem ..
but i can't figure it out and it is bugging the crap out of me ...
well here it is ...


How many different solutions to the equation X1+X2+X3+X4+X5+X6 = 149
ARE THERE IF THE VALUE OF EACH VARIABLE x1, x2, x3, x4, x5, x6 must be either 0 or 1 .


I think the answer is 0 or impossible .. but there might be some trick involve in it..
and that's wat been bugging me for half hour .....ARGGHhh

wat do you think ?

 

[Moonbeam]

I suppose you should do it discreetly.

 

[TurtleMan]

haha i tried... untill now i have to go public...

 

[rob3rt]

I don't understand what the problem is asking for? I can only choose 1 or 0 for each variable? Then isn't it just plain impossible since 1 + 1+ 1+ 1+ 1+ 1 = 6? How can it ever be 149?

 

[SuperTool]

Originally posted by: rob3rt
I don't understand what the problem is asking for? I can only choose 1 or 0 for each variable? Then isn't it just plain impossible since 1 + 1+ 1+ 1+ 1+ 1 = 6? How can it ever be 149?

but it could be 149 mod 143

 

[TurtleMan]

Originally posted by: SuperTool

Originally posted by: rob3rt
I don't understand what the problem is asking for? I can only choose 1 or 0 for each variable? Then isn't it just plain impossible since 1 + 1+ 1+ 1+ 1+ 1 = 6? How can it ever be 149?

but it could be 149 mod 143

u think it probalby mean 0 <= x <= 1 ?
shoo.. i dont know .. that's why it is confusing me ..

 

[hamburglar]

it looks impossible.

 

[royaldank]

Just guessing, but I guess you could start by proving 1 > 0, then put in 1 for everything (creating a maximum) and show it is < 149.

 

[aux]

As most people here said, the equation (as formulated) has no solution.
Check the equation again, you may have copied it incorrectly, for example it may be (as SuperTool suggested) modulo something.
Of course it may just be an easy equation with no solutions

 

[TurtleMan]

umm nope, that's exactly wat it say on the paper =]

this problem is from Generalized Permutations and Combinations

and the equation that i think would apply for it is just a simple C(n+r-1, r)

for a simple example, it would be like ..

X1 + X2 + X3 = 11

where x1,x2,x3 are nonnegative integers `


and then it would be C (3 + 11 - 1 , 11) <-- and it give you 78 solutions..

 

[CPA]

Is this question out of the book or given by the professor? I would check that it is not in error first.

 

[TurtleMan]

well it is given on the test... so it is correct =D