Probability that each man has a woman paired in a tennis game. 
6 men and 6 women are sorted randomly for doubles games in three tennis courts. What is the probability that each man is paired with a woman?

The provided answer stated that the distributions amongst the courts is irrelevant. The first man has 6 women to choose from. The second man has 5 women to choose from etc, which forms $6!$ ways.
For the total, the first person chooses one partner out of 11, then the pair leave. The second person chooses one partner out of 9, then the pair leave etc. This forms $11 \times 9 \times 7 \times 5 \times 3 \times 1$.
Hence, the probability is $$\frac{6!}{11 \times 9 \times 7 \times 5 \times 3 \times 1} = \frac{16}{231}$$
Problem:
It is not clear to me how the distributions amongst the courts is irrelevant. It seems to me that the solution above is more suitable for a question like 

6 men and 6 women are to be placed in 6 pairs randomly. What is the probability that each man is paired with a woman?

Surely including the tennis courts changes things because having the pair $(M_1 F_1)(M_2 F_2)$ in the same court against each other is the same as $(M_2 F_2)(M_1 F_1)$.
 A: No, the probabilities of both scenarios are the same. Recall that if $S$ is a sample space and $E\in S$, then
$$P(E)=\frac{\operatorname{elements in E}}{\operatorname{elements in S}}$$
Let $E$ be the event that each man is paired with a woman. If you take into account the sorting of pairs into courts, then the number of elements in the sample space is multiplied by $6$ (if you count the three courts as the same) since there are $6$ ways to pair up $6$ pairs. However, the number of elements in $E$ is also multiplied by $6$ when you do this for the same reason. Since we are taking a ratio of $\operatorname{length}(E)$ to $\operatorname{length}(S)$, the $\times 6$ will actually cancel out and we will be left with the same probability.
TLDR: The sorting of  pairs into tennis courts is irrelevant, because it does not change what proportion of scenarios contain man-woman pairs - it just adds more scenarios.
A: Another approach. We have 6 courts with a first player and second player each.
For the $(7-i)$-th court we can choose a man in $i$ ways and a woman in $i$ ways and each man can be first player or second player. So the probability that each man is paired with a woman is
$$\frac{\prod_{i=1}^6 (i\cdot i\cdot 2)}{12!}=\frac{6!\cdot 6!\cdot 2^6}{12!}=\frac{16}{231}.$$
