1
$\begingroup$

How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?

So I know there is a formula for computing the number of nonnegative solutions

${8+3-1 \choose 3-1}={10\choose 2}$

So I then just subtracted cases where one or two integers are $0$.

If just $x=0$ then there are $6$ solutions where neither $y,z=0$.

So I multiplied this by $3$, then added the cases where two integers are $0$

$3\cdot 6+3=21$. So I get ${10 \choose 2}+21=66$

For the last problem where $x,y,z\geq -3$ I'm not sure how to deal with it.

$\endgroup$
  • $\begingroup$ I would try $-3\le x,y,z\le0$, then add it to the solutions you already have. $\endgroup$ – R. Burton Feb 12 at 22:57
2
$\begingroup$

Note you can use the same method for the first question: if $x+y+z=8$ and $x,y,z>0 $, you can set $x=1+x', y=1+y', z=1+z'$, where $x',y',z'$ are nonnegative and $x'+y'+z'=5$.

Therefore the number of positive solutions is $\;\dbinom{5+3-1}{3-1}=\dbinom 72=21.$

More generally, it is easy to prove that the number of positive integer solutions of the equation $$x_1+x_2+\dots +x_r=n \quad (n\ge r)$$ is equal to $\;\dbinom{n-1}{r-1}$.

$\endgroup$
  • $\begingroup$ @N.F.Taussig: One never rereads oneself carefully enough… Fortunately, others do it for you! Thank you for pointing it! $\endgroup$ – Bernard Feb 12 at 23:18
  • $\begingroup$ What solutions am I missing here then? The pairs I should have where one integer is $0$ is $(1,7),(2,6),(3,5),(4,4)$ which gives me $8$ solutions where only one integer is $0$. Multiply that by $3$ for each case, $24$. Then there are three remaining solutions where two are $0$ and one is $8$. Gives me $27$ solutions which have a nonnegative. $45-27=18$. $\endgroup$ – AColoredReptile Feb 12 at 23:41
  • $\begingroup$ You have $7$ solutions when only one integer is $0$, plus the three solutions with two integers equal to $0$, so you have to subtract $3\times 7+2=24$. Furthermore, in your post, you added instead of subtracting. $\endgroup$ – Bernard Feb 12 at 23:51
1
$\begingroup$

For the second problem, write $x=-3+x'$ and so on. You have $x'+y'+z'=17$ and $x',\dots$ are nonnegative, a case you know how to solve.

You can also solve the first problem this way; now you would set $x=1+x'$, etc.

$\endgroup$
0
$\begingroup$

Start by looking at the number of solutions of \begin{eqnarray*} X+Y+Z=17 \end{eqnarray*} which has $\binom{17+3-1}{3-1}$ solutions.

This is the same as the number of solutions of \begin{eqnarray*} x+y+z=8 \end{eqnarray*} where $x=X-3$,$y=Y-3$ and $z=Z-3$.

$\endgroup$
0
$\begingroup$

Your answer to the question of how many solutions the equation $$x + y + z = 8 \tag{1} $$ has in the positive integers is incorrect. As a sanity check, observe that there must be fewer solutions to the equation in the positive integers than there are in the nonnegative integers since we are not allowed to substitute $0$ for any of the variables.

How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z > 0$?

If $x, y, z$ are positive integers, then $x' = x - 1$, $y' = y - 1$, and $z' = z - 1$ are nonnegative integers. Substituting $x' + 1$ for $x$, $y' + 1$ for $y$, and $z' + 1$ for $z$ in equation 1 yields \begin{align*} x' + 1 + y' + 1 + z' + 1 & = 8\\ x' + y' + z' & = 5 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers with $$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$ solutions. Notice that there are fewer solutions to equation 1 in the positive integers than the nonnegative integers, as we would expect.

How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z \geq -3$?

If $x, y, z \geq -3$, then $x' = x + 3$, $y' = y + 3$, and $z' = z + 3$ are nonnegative integers. Substitute $x' + 3$ for $x$, $y' + 3$ for $y$, and $z' + 3$ for $z$ in equation 1, then proceed as above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.