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The total number of positive integer solutions of $$21 \leq a + b + c \leq 25$$ is __________.
Correct Answer: 1160
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
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