Am I on the right track to figure out this combinatorics problem?

Question

Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?

My thought: r = red b = blue g = green y = yellow * = undetermined

Every arrangments should in this form: rrbbggyy**

Case 1: two * are same

$\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$

Case 2: two * are different

$\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$

Result = Case 1 + Case 2

Any suggestions would be appreciated!

Answer 1

It's two cases.

First. We have one quadruplet of 4 balls of the same color. There are 4 variations here to form quadruplets. The rest 6 balls will automatically form 3 doublets (to satisfy the conditions of the problem). The overall number of placing them into a sequence is permutations with repetition: $$\frac{10!}{4!\,2!\,2!\,2!}$$ But we have to multiply it by 4 because we have 4 variations. So it is $$4\cdot\frac{10!}{4!\,2!\,2!\,2!}$$

Second case. We have two triplets and two doublets. How many variations in terms of triplets and doublets? The answer is four choose two -- that is combinations. The rest is the same as above: $${4\choose2}\cdot\frac{10!}{3!\,3!\,2!\,2!}$$

And finally the overall number of possible strings of balls: $$4\cdot\frac{10!}{4!\,2!\,2!\,2!}+{4\choose2}\cdot\frac{10!}{3!\,3!\,2!\,2!}$$

Answer 2

The partition by distinct colors can be either $10=4+2+2+2$ or $10=3+3+2+2$. In the first case, you also need to choose the $1$ color that is to be used 4 times, and in the second case, the $2$ colors that are to be used $3$ times. Thus, the total number of choices is $$ \binom{4}{1}\binom{10}{4,2,2,2}+\binom{4}{2}\binom{10}{3,3,2,2} $$