The total number of positive integer solutions of $$21 \leq a + b + c \leq 25$$ is __________.

We are asking for positive integral solutions. So, each of the a, b, and c should be at least 1.

Let's assume a = x+1, b = y+1, c = z+1. Here, x, y and z can be zero.

The possible cases

I) a+b+c = 21

x+y+z = 18

The number of ways the above equation can be satisfied is $$\left(n+r-1\right)C_{r-1}$$ = $$\left(18+3-1\right)C_{3-1}=20C_2=190$$

II) a+b+c = 22

x+y+z = 19

The number of ways the above equation can be satisfied is $$\left(n+r-1\right)C_{r-1}$$ = $$\left(19+3-1\right)C_{3-1}=21C_2=210$$

III) a+b+c = 23

x+y+z = 20

The number of ways the above equation can be satisfied is $$\left(n+r-1\right)C_{r-1}$$ = $$\left(20+3-1\right)C_{3-1}=22C_2=231$$

IV) a+b+c = 24

x+y+z = 21

The number of ways the above equation can be satisfied is $$\left(n+r-1\right)C_{r-1}$$ = $$\left(21+3-1\right)C_{3-1}=23C_2=253$$

VI) a+b+c = 25

x+y+z = 22

The number of ways the above equation can be satisfied is $$\left(n+r-1\right)C_{r-1}$$ = $$\left(22+3-1\right)C_{3-1}=24C_2=276$$

The sum of all these numbers = 190+210+231+253+276 = 1160