I asked this question today. I have problem in the second part:
In how many ways can 4 red balls and 7 blue balls be arranged in 3 boxes where each box must contain at least 1 red ball and each box can contain less than or equal or 4 balls, under the following cases?
(1) Each ball is distinct and Each box is distinct
(2) Each ball is distinct and Each box is identical
I have calculated the first part using generating functions and another user gave me another method. However, we could not agree with the part $2$.
We know that the distribution of ball would be $(r^2b)(rb^3)(rb^3)$ or $(r^2b^2)(rb^2)(rb^3)$
- for $(r^2b)(rb^3)(rb^3)$:Let $(r^2b)=a,(rb^3)=b$ , we can arrange them in a line $3!/2!=3$ ways. Thus, $$3\times\binom42\binom71\binom21\binom63$$ if the boxes were distinct
If we want to make the boxes identical, then divide by $3!$, so $$\frac{1}{2}\times\binom42\binom71\binom21\binom63=840$$
- for $(r^2b^2)(rb^2)(rb^3)$ . Let $(r^2b^2)=a,(rb^2)=b,(rb^3)=c$ . We can arrange them in line by $3!$ ways. So, the answer is $$6\times\binom42\binom72\binom21\binom52$$ if the boxes were distinct.
If we want to convert them into identical, then divide by $3!$, so the answer is $$\binom42\binom72\binom21\binom52=2520$$
So, $$840+2520=3360$$
@trueblueanil claims that the answer is $4200$. Who is right ?