# Thread: # of integer solutions

1. ## # of integer solutions

Here's an example problem:
Find the number of integer solutions to the equation:
x1 + x2 + x3 + x4 = 20
with the restrictions:
x1 >=1; x2 >=3; x3 >= -3; x4 >= 0

To do this we create new variables as follows:

y1 = x1 - 1 >=0; y2 = x2 - 3 >=0; y3 = x3 + 3 >=0; y4 = x4 - 0 >=0

y1 + y2 + y3 + y4 = (x1-1) + (x2-3) + (x3+3) + (x4-0) = x1 + x2 + x3 + x4 -1 = 20 -1 = 19.

Thus, 19+4-1 C 4-1 = 22C3 integer solutions.

How do I go about it if my restrictions are as follows: (less than instead of greater)
x1 <= 8; x2 <=10; x3 <= 12; x4 <=5

Thansk

2. use the same method as u have done as above choosing y1= 8-x1 ad similarly y2,y3,y4

at the end then u will have y1+y2+y3+y4=35-20=15

it has integer solutions 15+4-1C4-1 = 15C3

3. Thanks!! Now, what would I do differently if there were NO restrictions on x1,...,x4?

4. Originally Posted by jzellt Thanks!! Now, what would I do differently if there were NO restrictions on x1,...,x4?
well it would be obvious that there can be infinitely many such solutions in such a case

infact the only time when u can give the number of silutions is when u have restraints on all the 'n' (out of the 'n' numbers being added) being restricted or at the most '(n-1)' numbers (out of the n numbers being added) being restricted

even if they restrict '(n-2)' numbers out of the n numbers, then u can always choose the remaining 2 numbers in infinitely many ways to get the desired result

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