Distributing 7 red balls and 2 blue balls into 3 containers such that in each container, there are 3 balls and at least 2 of them are red balls. How many ways are there to distribute 7 distinguishable red balls and 2 distinguishable blue balls into 3 indistinguishable containers such that in each container, there are 3 balls and at least 2 of them are red balls?

I got $\dfrac{\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 3\cdot 2}{3!}=630$, because there are $\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}$ to distribute 6 red balls so each container has 2 balls. Next, there are $3$ containers for the last 7 ball and 2 ways to distribute the 2 blue balls. Lastly, divide by $3!$ because the containers are indistinguishable. However, this is the wrong answer. Can someone tell me what's wrong with my solution?
Thanks!
 A: The only way they can be distributed to meet the conditions is:
$red\,:\;3-2-2$
$blue:\;0-1-1$
We need to compute separately for each color and multiply
thus $\Large\left(\frac{7!}{3!2!2!}\cdot\frac{1}{2!}\right) \times \left(\frac{2!}{0!1!1!}\cdot\frac{1!}{2!}\right)=210$
Explanation
Had they been distinct boxes, the number would have been
$\Large\left(\frac{7!}{3!2!2!}\cdot\frac{3!}{2!}\right) \times \left(\frac{2!}{0!1!1!}\cdot\frac{3!}{2!}\right)$
but as they are identical, we have replaced the numerators of the second term for each color by $1$

An obsevation
I find that some answers/answers in comments have been obtained with very little computation, which is all to the good, but I am a fan of the multinomial coefficient, and that is because the process is so mechanical that it can be done even in half-sleep.

*

*For putting distinct balls into distinct boxes, you just need to multiply two multinomial coefficients for $[Lay\;down\;pattern]\times[Permute\; pattern]$
eg if you roll a die $8$ times, and get a pattern $1-2-0-3-1-1$, the answer is
$\dbinom{8}{1,2,0,3,1,1}\dbinom{6}{3,1,1,1}$
Better simplified and understood as permutations (where $0!,1!$ etc can be conveniently removed), so
$\dfrac{8!}{2!3!}\times\dfrac{6!}{3!}$
And the only change required for distinct balls to identical boxes is to change the numerator of the second term to $1$, thus $\dfrac{8!}{2!3!}\times\dfrac{1}{3!}$
A: For another view point :
As @trueblueanil said , the only possible distribution can be done when one boxes has $3$ red balls and the other two take $2$ red balls. The blue balls will go directly to complement them to $3$ balls
To make our calculation easier , think that the boxes are distinct in the beginning.
Step 1-) Determine which box will have $3$ red balls by $C(3,1)$.
Step 2-) Distribute red balls to $3$  distinct boxes such that one box has three ball , the others have two balls by $C(7,3)\times C(4,2) \times C(2,2)$
Step 3-) Distribute $2$ blue balls to those boxes which have $2$ red balls such that each box will take $1$ blue balls ,so we can do it by $C(2,1) \times C(1,1)$.
Now we found the number of distributing $7$ distinct red balls and $2$ distinct blue balls to $3$ distinct boxes such that one box has $3$ red balls and the other two boxes have $2$ red balls and $1$ blue ball. It is equal to $$C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)=1260$$
However, we know that those three boxes were identical , so divide the result by $3!$ to convert distinct boxes into identical.
$$\frac{C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)}{3!}=210$$
A: You want to count partitions in three sets of the set $\{r_1,\ldots ,r_7,b_1,b_2\}$ (there $r_k$ represents red balls, and $b_k$ blue balls) such that every set in the partition had exactly three balls and the blue balls cannot be together in one of the sets.
Note that these partitions are determined by just two sets each one with two red balls and one blue ball, as the third set will be determined by the other balls. Moreover: we can differentiate these two sets depending on the blue ball that contain, thus we have partitions determined by these two sets
$$
\{r_j,r_k,b_1\}\text{ and }\{r_l,r_i,b_2\}
$$
Thus the number of these sets are determined by pairs of distinct sets $(\{j,k\},\{l,i\})$ such that $j,k,l$ and $i$ are distinct elements of $I:=\{1,\ldots,7\}$. This is equivalent to, first, choose all distinct subsets of two elements of $I$, and after with the remaining elements choose over all distinct subsets of two elements.
This amounts to $\binom{7}{2}\cdot \binom{5}{2}=7\cdot 6\cdot 5=210$.
