Generalization of matching problem for knives and forks Inspired by Probability of knives and forks matching, the following problem occurs to me:
Assume we have $3n$ knives and $3n$ forks, of which $n$ knives and forks are red, $n$ knives and forks are black, and $n$ knives and forks are white.
Now assume the knives and forks are randomly paired with one another.  What is the probability that no knife is paired with a fork of the same color?
Note that for $n=1, 2$, the probability is $\frac{1}{3^n}$.  (I provided the answer to the $n=2$ case in the linked question.)  Is that the general answer?  If there are $k$ colors instead of $3$ colors, is the answer for the analogous problem $\frac{1}{k^n}$?
 A: I am getting that the probability is $$\frac{(n!)^3}{(3n)!} \cdot \sum_{a = 0}^n \binom{n}a^3.$$
Let $I_1 = 1 \cdots n, I_2 = n+1 \cdots 2n, I_3= 2n+1 \cdots 3n$. We can think of the desired matching as a permutation $f$ on $3n$ elements such that $f$ satisfies the following: $f$ applied to $I_j$ avoids $I_j$ for all $j$.
To construct such a $f$, we first specify the values that it takes for $I_1$. The number of ways to pick integers for these values is $\binom{n}a \binom{n}{n-a}$ for some $a \ge 0$ since we need to pick $a$ integers from $n+1$ to $2n$ and $n-a$ from $2n+1$ to $3n$. 
Now this means that all the left over integers from $2n+1$ to $3n$ that we didn't pick as the image of some element in $I_1$ must be the image of some element in $I_2$. There are $a$ such left over integers. Then we need to pick $n-a$ integers from $1$ to $n$ since the range of $f$ applied $I_2$ needs $n$ integers. This completely determines the range for $f$ applied to $I_3$.
Finally, for each $I_j$, we can permute the values that they are mapped to and still satisfy our desired condition. There are $(n!)^3$ such ways to do this. Dividing by the total number of permutations gives us the probability. 
The generalization to $k$ buckets is clear but we instead get multinomial sums. Hopefully some nice asymptotic is possible. 
