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Question Number 116881 by bemath last updated on 07/Oct/20
Find the number m of non negative  integer solution of x+y+z=18
Findthenumbermofnonnegativeintegersolutionofx+y+z=18
Answered by john santu last updated on 07/Oct/20
we can view each solution ,say x=3, y=7 ,z=8  as a combination of r = 18 objects  consisting of 3 a′s , 7 b′s and 8 c′s, where   there are M = 3 kinds of objects a′s, b′s   and c′s . Then m = C(r+M−1,M−1)   = C(20,2) = ((20×19)/(2×1)) = 190
wecanvieweachsolution,sayx=3,y=7,z=8asacombinationofr=18objectsconsistingof3as,7bsand8cs,wherethereareM=3kindsofobjectsas,bsandcs.Thenm=C(r+M1,M1)=C(20,2)=20×192×1=190
Answered by mr W last updated on 07/Oct/20
(1+x+x^2 +x^3 +...)^3   =(1/((1−x)^3 ))=Σ_(k=0) ^∞ C_2 ^(k+2) x^k   k=18:  C_2 ^(20) =((20×19)/2)=190
(1+x+x2+x3+)3=1(1x)3=k=0C2k+2xkk=18:C220=20×192=190
Answered by mr W last updated on 07/Oct/20
x+y=n  x=0,1,...,n ⇒n+1 solutions    n=18−z=0..18    ⇒totally 1+2+...+19=((19×20)/2)=190
x+y=nx=0,1,,nn+1solutionsn=18z=0..18totally1+2++19=19×202=190

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