Here's the question you clicked on:
giovkast
determine the number of solutions for x part natural numbers
\[{x1}, ...., {x6} = 15\]
\[\left(\begin{matrix}15 + 6 - 1 \\ 15\end{matrix}\right)\] is this the answer?
what kind operations there, add, subtrac, or multiplication or division ?
is it like a+b+c+d+e+f=15 ? with a,b,c,d,e, and f are natural numbers
discrete math, ow sorry its addition
hehe... i have guessed actually, i will use the formula : |dw:1355741391689:dw|
why did you use (15 - 1)?
wait... looks i have mistaken
what did you do wrong then, exactly?
ok, for the first... i have mistaken after i do some experiments, it shoulde be |dw:1355743741310:dw|
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=5-b 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=4-b 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=2-b 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(6-1, 3-1) = C(5,2) = 10
there are 6 x's, from x1 till x6 and together they add up to 15
with same idea, if given the problem : a+b+c+d=8 so, the total ways = C(8-1,4-1) = C(7,3) = 35 ways btw, i have done it by manual also like the first ^ (the answer is same)
yeah, like i said before it will be C(15-1,6-1) = C(14,5) = 14!/(9!5!) = ....
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 ......
thank you very much for the explanation