I want to find the number of solutions to the following equation $$x_1+x_2+x_3+x_4 = 21$$ such that $x_i$ is in $[0,10]$

I know that total solutions will be ${24}\choose3$ but this will also include solutions where $x_i$ is not in $[0,10]$. So how do I find the solution to this problem?

  • $\begingroup$ The quantity $\binom{20}{3}$ is actually the number of solutions in the positive integers, not the nonnegative integers. $\endgroup$ – N. F. Taussig Dec 22 '18 at 13:51
  • $\begingroup$ Yeah right, I’ve made the correction, thanks $\endgroup$ – user601297 Dec 22 '18 at 13:52

As you observed, the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 21 \tag{1}$$ in the nonnegative integers is $$\binom{21 + 4 - 1}{4 - 1} = \binom{24}{3}$$ From these, we must subtract those solutions in which one or more of the variables exceeds $10$. There can be at most one such variable since $2 \cdot 11 = 22 > 21$.

We count the number of solutions that violate the restriction that $x_i \leq 10$, $1 \leq i \leq 4$.

Choose which of the four variables exceeds $10$. Suppose it is $x_1$. Then $x_1' = x_1 - 11$ is a nonnegative integer. Substituting $x_1' + 11$ for $x_1$ in equation 1 yields \begin{align*} x_1' + 11 + x_2 + x_3 + x_4 & = 21\\ x_1' + x_2 + x_3 + x_4 & = 10 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers with $$\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3}$$ solutions. Hence, there are $$\binom{4}{1}\binom{13}{3}$$ solutions in the nonnegative integers that violate the restriction that $x_i \leq 10$, $1 \leq i \leq 4$.

Hence, the number of admissible solutions is $$\binom{24}{3} - \binom{4}{1}\binom{13}{3}$$


I would find coefficient of $x^{21}$ in the expansion of $$(1+x+x^2+x^3+...+x^{10})^4$$ Calculate $1+x+x^2+x^3+...+x^{10}$ by G.P. and then using formulas for finding coefficients for binomial and negative binomial you can find the coefficient.

  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Dec 22 '18 at 14:03
  • $\begingroup$ @N.F.Taussig thank you ...i first tried typing but I typied $x^21$ so I got confused $\endgroup$ – onlymath Dec 22 '18 at 14:04

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.