In how many ways to NOT take a red ball in an urn?

Question

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.

Answer 1

If you don't need red balls, this boils down to a simpler problem: "in how many ways can I fetch 10 balls out of an urn with 20 unique balls in it". If order isn't important, this is simple:

$${20 \choose 10} = \frac{20!}{10!10!}$$

Answer 2

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.