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
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
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