Balls without being together (Exercise)

Question

I have the following problem, it's of my old maths contest, and i'd like to see how can i solve it because i'm not really good at this math area.

Exercise 2.- If you have 5 blue balls, 5 red balls and 5 white balls, How many ways can you order them without blue balls together?

I did this:

$P_n^{5,5,5}$ - $P_{n-4}^{5-4,5,5}$ = $\frac{15!}{5!5!5!}$ - $\frac{11!}{1!5!5!}$

But i don't think it's correct, i don't even have more ideas. I'd like to do it by a binomial coefficient or by any permutation/variation formula. I would appreciate for any help, thanks.

Answer 1

Without all $5$ blue balls together:

$$\frac{(5+5+5)!}{5!\times5!\times5!}-\frac{(5+5+1)!}{5!\times5!\times1!}$$

---

Without any $2$ blue balls together:

$$\frac{(5+5)!}{5!\times5!}\times\binom{5+5+1}{5}$$