let $x_1$ be the number of green balls
let $x_2$ be the number of yellow balls
let $x_3$ be the number of blue balls
$x_1 + x_2 + x_3 = 10 ,$ $0 \leq x_1<11, \ \ 0 \leq x_2<6, \ \ \ 0 \leq x_3<4$
Total possible solutions without any restriction = $\binom{12}{2} = 66$
Now let's calculate number of invalid solutions,
Total solutions when $\color{red}{x_2 \geq 6,}$
$x_1 + x_2 + 6 + x_3 = 10$
$x_1 + x_2 + x_3 = 4$
$ = \binom{6}{2} = 15$
Total solutions when $\color{red}{x_3 \geq 4,}$
$x_1 + x_2 + x_3 + 4 = 10$
$x_1 + x_2 + x_3 = 6$
$ = \binom{8}{2} = 28$
Total solutions when both $\color{red}{x_3 \geq 4}$ and $\color{red}{x_2 \geq 6,}$
$x_1 + x_2+ 6 + x_3 + 4 = 10$
$x_1 + x_2 + x_3 = 0$
$ = \binom{2}{2} = 1$
Total invalid solutions $= 15 + 28 - 1 = 42$
Total valid solutions = $66 - \color{red}{42} = 24$