how many number of solutions to the equations a+b+c+d = 11 where 1<= a,b,c,d <=5?

12c3

a+b+c+d = 7 where 0 <= a,b,c,d <= 5

The number of solutions of this equation where there is no restriction on the upper limit of the variables is $$^{7+4-1}C_{4-1}$$ =$$^{10}C_3$$ = 120
From this we have to subtract the number of solutions in which any of the variables take a value more than 5

Let a = 6 => b+c+d = 1 => Number of solutions = $$^{1+3-1}C_{3-1}$$ = $$^3C_2$$ = 3 solutions
Let a = 7 => b+c+d = 0 => Only 1 solution
So, a total of 4 solutions have to be subtracted if 'a' takes a value more than 5. The variable that can take the value more than 5 can be selected from among a,b,c,d in 4 ways. So, a total of 4*4 = 16 solutions have to be subtracted from 120.

Answer = 120 - 16 = 104