# combination

• Jan 28th 2009, 01:00 AM
tukilala
combination
what is the number of solution of x+y+z=17 in non- negative integers is:
the book answer is: (17+3-1 C 17)=19C17=171
but i dont unserstand why 17 and not 18???
because there are 18 digits from 0 to 10
so the answer not sepose to be (18+3-1 C 17) = 20C17
???
• Jan 28th 2009, 03:08 AM
Plato
I explained it to you in this reply.
http://www.mathhelpforum.com/math-he...tion-help.html
Think of the '17' as 17 identical one to be put into the different variables.
• Jan 28th 2009, 04:13 AM
tukilala
but why k=18?
i understand the formula
but my question is that there are 18 digits from 0 to 17
so n sepose to be 18, no?
but the book show this solution: (17+3-1)C(17)
i dont understand why its not (18+3-1)C(18)
???
• Jan 28th 2009, 04:59 AM
Plato
Quote:

Originally Posted by tukilala
but why k=18?
i understand the formula
but my question is that there are 18 digits from 0 to 17
so n sepose to be 18, no?
but the book show this solution: (17+3-1)C(17)
i dont understand why its not (18+3-1)C(18)???

why its not? (18+3-1)C(18) because that is wrong!
We are placing 17 ones into three variables not 18.
Where are you getting 18 from? The problem is clearly about 17.
Why are you changing the numbers at will?
• Jan 28th 2009, 07:29 AM
tukilala
and if one of the variables is 0 ?
every variable have 18 possibility numbers :0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17

this is where i get the 18 from....

so whats can u explain me please again why not 18 numbers???
thanks
• Jan 28th 2009, 08:16 AM
Plato
You have only three variables: $x,\;y,\,\;\& \,z$.
Here is a list of very few of the possible solutions:
$\begin{array}{ccccc} x &\vline & y &\vline & z \\
\hline 1 &\vline & 7 &\vline & 9 \\ 9 &\vline & 1 &\vline & 7 \\
7 &\vline & 9 &\vline & 1 \\ 8 &\vline & 9 &\vline & 0 \\
0 &\vline & 0 &\vline & {17} \\ 0 &\vline & 1 &\vline & {16} \\
{15} &\vline & 1 &\vline & 1 \\ 4 &\vline & 5 &\vline & 8 \\
3 &\vline & {11} &\vline & 3 \\ \end{array}$
.
Think of it this way. You are putting 17 identical 1’s into three boxes labeled $x,\;y,\,\;\& \,z$.
Where is there an 18? Nowhere!