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

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

$x_1+x_2+x_3+x_4 = 17$

$x_1= x_i + 1$

$x_2= x_{ii}+2$

$x_3= x_{iii}+3$

$x_4= x_{iiii}$

So,

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

$x_i+x_{ii}+x_{iii}+x_{iiii}=11$

$3$ bars, $\binom{14}{3} =364$ different ways?
Is this the right approach? correct answer?
for the inequality x1+x2+x3&lt;=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.
for the inequality x1+x2+x3&lt;=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.