$[x^{25}] (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)$
$[x^{25}] (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^5$
$[x^{25}] x^5(x^0 + x^1 + x^2 + x^3 + x^4 + x^5)^5$
$[x^{20}] (x^0 + x^1 + x^2 + x^3 + x^4 + x^5)^5$
$[x^{20}] (1 + x^1 + x^2 + x^3 + x^4 + x^5)^5$
so now the equation is,
$x_1 + x_2 + x_3 + x_4 + x_5 = 20, (\ 0 \leq x_1 \leq 5,0 \ \leq x_2 \leq 5,0 \leq x_3 \leq 5,0 \leq x_4 \leq 5, \ 0 \leq x_5 \leq 5 )$
Total possible solutions = $\large \binom{24}{4}$
Total valid solutions = total solutions - invalid solutions
So we'll try to calculate all invalid solutions,
let $x_1 \geq 6$
$x_1 + 6 + x_2 + x_3 + x_4 + x_5 = 20$
$x_1 + x_2 + x_3 + x_4 + x_5 = 20 - 6$
$x_1 + x_2 + x_3 + x_4 + x_5 = 14$
so number of invalid solutions where $x_1 \geq 6$ = $\large \binom{18}{4}$
so the number of invalid solutions where any one of $x_i \geq 6 =$ $\large 5 \times \binom{18}{4} = $$15300$
Now we'll calculate the number of ways $x_1 \geq 6$ and $x_2 \geq 6$
$x_1 + 6 + x_2 + 6 + x_3 + x_4 + x_5 = 20$
$x_1 + x_2 + x_3 + x_4 + x_5 = 8$
so number of invalid solutions = $\large \binom{12}{4}$
but this was only when we put constrain on $x_1$ and $x_2$ but there can be $\binom{5}{2} = 10$ combinations.
So possible combinations = $10 \times \large \binom{12}{4} = 4950$
Now we'll calculate the number of ways $x_1 \geq 6$, $x_2 \geq 6$ and $x_3 \geq 6$,
$x_1 + 6 + x_2 + 6 + x_3 + 6 + x_4 + x_5 = 20$
$x_1 + x_2 + x_3 + x_4 + x_5 = 2$
similarly there are $\binom{5}{3} = 10$ combinations.
So possible combinations = $10 \times \large \binom{6}{4} = 150$
Now we cannot go further finding invalid solutions because 4 variable at once cannot $\geq 6$ all at once.
So total invalid solutions = $15300 - 4950 + 150 = 10500$
Total valid solutions = $\large \binom{24}{4} - $$13560 = \large \color{blue}{ 126}$