# How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if

How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if all $x$ are non-negative integer:

• No restrictions. The solution is $C(55 + 4, 4) = C(59,4)$ but I fail to see why, can someone explain this to me?
• Every $x_k$ is odd.
• If $x_1>=1,x_2>=2,x_3>=3,x_4>=2,x_5>=1,$
• A similar problem has been worked out here: cs.uiuc.edu/class/fa06/cs173/homework/solutions/hw10sol.pdf – user9413 May 22 '11 at 12:41
• Wow, thanks I'll have a read. – meiryo May 22 '11 at 13:16
• "No restrictions" would allow real numbers or complex numbers. So perhaps you mean: restricted to nonnegative integers... – GEdgar May 22 '11 at 13:19
• Yes, I'll have to fix that – meiryo May 22 '11 at 13:21

1. You forgot to say that $0\le x_i$ (otherwise there is an infinite number of solutions). This is a standard "choice with repetitions and without order" case (hence the formula, which is one of the basic formulas in combinatorics). In this case, you have 55 "balls" to distribute freely into the five "holes" - the variables.
2. Here you can write $x_i=2y_i+1$ and solve for $y_i$ instead of $x_i$. You'll get something like $2(y_1+\dots+y_5)+5=55$ - simplify to get $y_1+\dots+y_5=25$ (how?)
3. Here you use $x_1=y_1+1$, $x_2=y_2+2$ and then again solve for $y_i$.