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giovkast
Group Title
determine the number of solutions for x part natural numbers
 one year ago
 one year ago
giovkast Group Title
determine the number of solutions for x part natural numbers
 one year ago
 one year ago

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giovkast Group TitleBest ResponseYou've already chosen the best response.0
\[{x1}, ...., {x6} = 15\]
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
\[\left(\begin{matrix}15 + 6  1 \\ 15\end{matrix}\right)\] is this the answer?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
what kind operations there, add, subtrac, or multiplication or division ?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
is it like a+b+c+d+e+f=15 ? with a,b,c,d,e, and f are natural numbers
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
discrete math, ow sorry its addition
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
hehe... i have guessed actually, i will use the formula : dw:1355741391689:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
sorry if im wrong...
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
why did you use (15  1)?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
wait... looks i have mistaken
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
what did you do wrong then, exactly?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
ok, for the first... i have mistaken after i do some experiments, it shoulde be dw:1355743741310:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
let's discuss for simple example below : a+b+c=6, for a,b,c are natural numbers. # if a=1, then b+c=5 or c=5b to get c be integer , so satisfied for b=1,2,3,4 (there are 4 ways) # if a = 2, then b+c=4, or c=4b to get c be integer , so satisfied for b=1,2,3, (there are 3 ways) # if a = 3, then b+c=3, or c=2b to get c be integer , so satisfied for b=1,2, (there are 2 ways) # if a = 4, then b+c=2, satisfied just for b=1 and c=1 ( 1 way) so, the total ways = 4+3+2+1 = 10 it just similar by C(61, 31) = C(5,2) = 10
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
there are 6 x's, from x1 till x6 and together they add up to 15
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
with same idea, if given the problem : a+b+c+d=8 so, the total ways = C(81,41) = C(7,3) = 35 ways btw, i have done it by manual also like the first ^ (the answer is same)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
yeah, like i said before it will be C(151,61) = C(14,5) = 14!/(9!5!) = ....
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
but, this method will be different, if a,b,c,d,... are the whole numbers :) thanks for ur question, i become more knowing for this problems ......
 one year ago

giovkast Group TitleBest ResponseYou've already chosen the best response.0
thank you very much for the explanation
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
you're welcome :)
 one year ago
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