Your $N^N$ looks reasonable for the total number of equally likely arrangements.
One approach to the second part is to say that we choose $2$ of the $N$ items ${N\choose 2}$ ways and a drawer to put them in $N$ ways and another drawer to be empty $N-1$ ways and then organise the remaining $N-2$ drawers and $N-2$ items $(N-2)!$ ways, giving $${N\choose 2}N(N-1)\,(N-2)! = \frac{N(N-1)}2N!$$ possibilities, which we then divide by $N^N$ to get a probability.
- For $N=2$ this gives $\frac12$
- For $N=3$ this gives $\frac23$
- For $N=4$ this gives $\frac9{16}$
- For $N=5$ this gives $\frac{48}{125}$
which seems plausible
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 $\dfrac{8!}{2!3!}\times\dfrac{1}{3!}$
Best Answer
Your numerator is larger than your denominator, so your result cannot represent a probability. It is best to avoid choosing representatives of a set, then selecting additional members of a set since that leads to over counting.
Method 1: There are three choices for each marble, so there are $3^6$ ways to distribute the marbles.
For the favorable cases, we need to subtract those configurations in which there are one or more empty containers.
There are three ways to select a container to be left empty and $2^6$ ways to distribute the marbles to the remaining two containers.
However, if we subtract this amount from $3^6$, we will have subtracted each case in which there are two empty containers twice, once for each way of designating one of the empty containers as the empty containers. We only wish to subtract these three cases once, so we must add them to the total. Hence, the number of favorable cases is $$3^6 - \binom{3}{1}2^6 + \binom{3}{2}1^6$$ by the Inclusion-Exclusion Principle.
Therefore, the probability that all containers are non-empty is $$\Pr(\text{all containers are non-empty}) = \frac{3^6 - \dbinom{3}{1}2^6 + \dbinom{3}{2}1^6}{3^6}$$
Method 2: Alternatively, observe that there are three ways to express $6$ as a sum of exactly three positive integers: \begin{align*} 6 & = 4 + 1 + 1\\ & = 3 + 2 + 1\\ & = 2 + 2 + 2 \end{align*}
Four marbles in one container, with one each in the other two containers: There are three ways to select the container which receives four marbles, $\binom{6}{4}$ ways to select which four of the six marbles are placed in that container, and two ways to distribute the remaining marbles to the remaining two containers. Hence, there are $$\binom{3}{1}\binom{6}{4}\binom{2}{1}\binom{1}{1}$$ such distributions.
Three marbles in one container, two marbles in another container, and one marble in the other container: There are three ways to select the container which receives three marbles, $\binom{6}{3}$ ways to select which three of the six marbles that container receives, two ways to select the container which receives two marbles, $\binom{3}{2}$ ways to select which two of the remaining three marbles are placed in that container, and one way to place the remaining marble in the remaining container. Hence, there are $$\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}\binom{1}{1} = 3!\binom{6}{3}\binom{2}{1}\binom{1}{1}$$ such distributions.
Two marbles are placed in each of the three containers: There are $\binom{6}{2}$ ways to place two of the six marbles in the first container, $\binom{4}{2}$ ways to place two of the remaining four marbles in the second container, and one way to place both of the remaining marbles in the third container. Hence, there are $$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ such configuration.
Hence, $$\Pr(\text{all containers are non-empty}) = \frac{\dbinom{3}{1}\dbinom{6}{4}\dbinom{2}{1}\dbinom{1}{1} + 3!\dbinom{6}{3}\dbinom{3}{2}\dbinom{1}{1} + \dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}{3^6}$$