How many integer solutions are there to the inequality $x_1 + x_2 + x_3 \leq 17$, if we require that $x_1 \geq 1$, $x_2 \geq 2$, $x_3 \geq 3$?

The way I did it:

$x_4 = 17-x_1-x_2-x_3-x_4$

$x_1+x_2+x_3+x_4 = 17$

$x_1= x_i + 1$

$x_2= x_{ii}+2$

$x_3= x_{iii}+3$

$x_4= x_{iiii}$


$x_i+1+x_{ii}+2+x_{iii}+3+x_{iiii}= 17$


$3$ bars, $\binom{14}{3} =364$ different ways? Is this the right approach? correct answer?

  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly. $\endgroup$ – N. F. Taussig Jul 5 '18 at 1:22
  • $\begingroup$ If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions. $\endgroup$ – fanvacoolt Jul 5 '18 at 2:49
  • $\begingroup$ Try summation over generating function. $\endgroup$ – linear_combinatori_probabi Jul 5 '18 at 4:27
  • 1
    $\begingroup$ Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $\checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer. $\endgroup$ – amWhy Jul 6 '18 at 23:01

for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.


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