How many integer solutions satisfy x1 + x2 + x3 = 20 with x1 2, x2 3, x3 4?

Substitute. Let y1=x1-2,, y2=x2-3,, y3=x3-4, each 0. Then y1+y2+y3 = 20-9 = 11. Stars and bars. Number of nonnegative solutions is 11+3-13-1 = 132.

Bounded Integer Solutions

Combinatorics

Integer Solutions

Easy

How many integer solutions satisfy

x_{1} + x_{2} + x_{3} = 20

with

x_{1} \geq 2

x_{2} \geq 3

x_{3} \geq 4

Hint 1

Substitute $y_{i} = x_{i} - minimum$ to reduce to a nonnegative stars-and-bars problem.

Hint 2

The total after substitution should sum to 11.

Show worked solution

1. Substitute
Let $y_{1} = x_{1} - 2, y_{2} = x_{2} - 3, y_{3} = x_{3} - 4$ , each $\geq 0$ . Then $y_{1} + y_{2} + y_{3} = 20 - 9 = 11$ .
2. Stars and bars
Number of nonnegative solutions is $(3 - 1 11 + 3 - 1) = (2 13)$ .
$(2 13) = 78$

Answer: 78