In how many ways to NOT take a red ball in an urn? There are 25 balls in five different colors in an urn. The balls are numbered 1 to 5 in each color. I.e. they are all unique.
In how many ways can I take 10 balls without returning any of them and I DON'T get any red ball.
Maybe I make this harder than it is. I've trying to solve it by inclusion-exclusoin and generating functions but I don't really get any reasonable answer.
By generating functions i tried to solve that I take at least one red ball and subtract that from $25 \choose 10$ that is the total ways of choosing 10 balls.
My first try was something like this where every x represent the different colors and $x_1$ is the red.
$x_1 + x_2 + x_3 + x_4 + x_5 = 10$
$1 \leq x_1 \leq 5, \\  0 \leq x_i \leq 5 , \\ 2 \leq i \leq 5$
From this we get
$5 + 9 - 1 \choose 9$ = 715.
I didnt like that answer so I tried something else.
$x(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)^4=10$
And from that, by $\frac{1-x^{n+1}}{1-x}$ we get
$x \cdot (\frac{1-x^5}{1-x})(\frac{1-x^6}{1-x})^{^4} = 10 $
calculating the coefficient of 10 we get 505 and that doesn't either feel right.
By inclusion - exclusion:
${25 \choose 10} - {24 \choose 9} + {23 \choose 8} - + - \dots $ But here I dont take respect to the red balls in any way.
My thought is to perhaps divide the problem into different cases, like
I take one red ball 
I take two red balls
$\vdots$
And make something from that but now I am pretty confused generally.

If you do not select any red balls, then you must select $10$ of the remaining $25 - 5 = 20$ balls, which can be done in
$$\binom{20}{10}$$
ways.
As a sanity check, observe that the number of ways of selecting exactly $k$ of the five red balls and $10 - k$ of the remaining $20$ balls is
$$\binom{5}{k}\binom{20}{10 - k}$$
Hence, the total number of selections is
$$\binom{5}{0}\binom{20}{10} + \binom{5}{1}\binom{20}{9} + \binom{5}{2}\binom{20}{8} + \binom{5}{3}\binom{20}{7} + \binom{5}{4}\binom{20}{6} + \binom{5}{5}\binom{20}{5} = \binom{25}{5}$$
which can be checked by direct calculation or by applying Vandermonde's identity.
