anonymous
  • anonymous
determine the number of solutions for x part natural numbers
Discrete Math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[{x1}, ...., {x6} = 15\]
anonymous
  • anonymous
\[\left(\begin{matrix}15 + 6 - 1 \\ 15\end{matrix}\right)\] is this the answer?
RadEn
  • RadEn
what kind operations there, add, subtrac, or multiplication or division ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

RadEn
  • RadEn
is it like a+b+c+d+e+f=15 ? with a,b,c,d,e, and f are natural numbers
RadEn
  • RadEn
hello ?
anonymous
  • anonymous
discrete math, ow sorry its addition
RadEn
  • RadEn
hehe... i have guessed actually, i will use the formula : |dw:1355741391689:dw|
RadEn
  • RadEn
sorry if im wrong...
anonymous
  • anonymous
why did you use (15 - 1)?
RadEn
  • RadEn
wait... looks i have mistaken
anonymous
  • anonymous
what did you do wrong then, exactly?
RadEn
  • RadEn
ok, for the first... i have mistaken after i do some experiments, it shoulde be |dw:1355743741310:dw|
RadEn
  • RadEn
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
anonymous
  • anonymous
there are 6 x's, from x1 till x6 and together they add up to 15
RadEn
  • RadEn
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)
RadEn
  • RadEn
yeah, like i said before it will be C(15-1,6-1) = C(14,5) = 14!/(9!5!) = ....
RadEn
  • RadEn
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 ......
anonymous
  • anonymous
thank you very much for the explanation
RadEn
  • RadEn
you're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.