8
$\begingroup$

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 $x_1\ge1,x_2\ge2,x_3\ge3,x_4\ge2,x_5\ge1$
$\endgroup$
4
  • 2
    $\begingroup$ A similar problem has been worked out here: cs.uiuc.edu/class/fa06/cs173/homework/solutions/hw10sol.pdf $\endgroup$
    – user9413
    May 22, 2011 at 12:41
  • $\begingroup$ Wow, thanks I'll have a read. $\endgroup$
    – meiryo
    May 22, 2011 at 13:16
  • 2
    $\begingroup$ "No restrictions" would allow real numbers or complex numbers. So perhaps you mean: restricted to nonnegative integers... $\endgroup$
    – GEdgar
    May 22, 2011 at 13:19
  • $\begingroup$ Yes, I'll have to fix that $\endgroup$
    – meiryo
    May 22, 2011 at 13:21

1 Answer 1

Reset to default
11
$\begingroup$
  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 $y_i$.
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .