Counting number of integer solutions

Question

How many integer solutions are there to the equation $a + b + c = 21$ if $a \geq3$, $b \geq 1$, and $c \geq 2b$?

Answer 1

If $a = 3+A$, $b =1+B$, $c = 2b + d=2B+d+2$, the equation is $A + 3 B + d = 15$ with $A, B, d \ge 0$. Now for each $B$ from $0$ to $5$, the number of $(A,d)$ pairs is $16 - 3 B$. So...

Answer 2

$a$ can vary from $3$ to $18$ $b$ can vary from $1$ to $16$ $c$ can vary from $2$ to $17$ Hence according to me number of solutions should be $16*16*16=4096$

Answer 3

Brute forcing: $$\begin{align}(a,b,c)=&(3,1,17), (3,2,16), (3,3,15), (3,4,14), (3,5,13), (3,6,12),\\ &(4,1,16), (4,2,15), (4,3,14), (4,4,13), (4,5,12),\\ &(5,1,15), (5,2,14), (5,3,13), (5,4,12), (5,5,11),\\ &(6,1,14), (6,2,13), (6,3,12), (6,4,11),\\ &(7,1,13), (7,2,12), (7,3,11), (7,4,10),\\ &(8,1,12), (8,2,11), (8,3,10), (8,4,9),\\ &(9,1,11), (9,2,10), (9,3,9), (9,4,8),\\ &(10,1,10), (10,2,9), (10,3,8),\\ &(11,1,9), (11,2,8), (11,3,7),\\ &(12,1,8), (12,2,7), (12,3,6),\\ &(13,1,7), (13,2,6),\\ &(14,1,6), (14,2,5),\\ &(15,1,5), (15,2,4),\\ &(16,1,4),\\ &(17,1,3),\\ &(18,1,2). \end{align}$$ Counting: $$6+2\cdot 5+4\cdot 4+3\cdot 3+3\cdot 2+3\cdot 1=50.$$