Probability of compatible love languages under uniform sampling In Chapman's five love languages he claims that everyone has a primary and a secondary love language.
Two people are said to be 'compatible' if their first love languages match, or the first love language of each partner matches the second love language of the other.
By counting I found that the fraction of conceivable couples matching this description is $\frac{60}{119}$, which is quite close to 50%. I counted with the following code:
from itertools import permutations
perms = [i for i in permutations(range(1,6), 5)]
count = 0
for i in perms:
    for j in perms:
        if i[0] == j[0] or (i[0] == j[1] and i[1] == j[0]):
            count += 1
print(count)

I further reasoned that there were ${5!\choose 2}$ possible pairings, thus giving us a ratio of 3600 / 7140.
Is there a symbolic calculation that better motivates the value of this ratio?
 A: Your calculation appears to assume that the members of the couple cannot have the same love-language-ordering, which is not part of the specification of the problem.   I think you have also calculated the numerator incorrectly.  In any case, I'll show you how you can do this problem.
It is possible to obtain the relevant combinations using appropriate theory on permutations.  However, rather than trying to count the combinations, you can proceed more easily by directly computing the probabilities from the relevant events.  Let $\mathscr{E}_1$ denote the event that the couple have the same primary-language and let $\mathscr{E}_2$ denote the event that the couple have a different primary-language but have the required primary-secondary match.  Under the specification given the couple are "compatible" if the event $\mathscr{E}_1 \cup \mathscr{E}_2$ occurs.  Assuming that all language-orderings are equally likely for each member of the couple (and their language-orderings are independent), the probability they are compatible is:
$$\begin{align}
\mathbb{P}(\text{Compatible}) 
&= \mathbb{P}(\mathscr{E}_1 \cup \mathscr{E}_2) \\[6pt]
&= \mathbb{P}(\mathscr{E}_1) + \mathbb{P}(\bar{\mathscr{E}}_1 \cap \mathscr{E}_2) \\[6pt]
&= \mathbb{P}(\mathscr{E}_1) + \mathbb{P}(\bar{\mathscr{E}}_1) \times \mathbb{P}(\mathscr{E}_2|\bar{\mathscr{E}}_1) \\[6pt]
&= \frac{1}{5} + \frac{4}{5} \times \frac{1}{4} \cdot \frac{1}{4} \\[6pt]
&= \frac{1}{5} + \frac{1}{20} \\[6pt]
&= \frac{1}{4}. \\[6pt]
\end{align}$$
This can be confirmed using a simple calculation in R.  Without loss of generality, suppose that the language-ordering of the first person is 1-2-3-4-5 and look over all possible orderings of the second person (assumed to be equally likely).  As can be seen, this confirms the above probability.
#Generate list of all possible language-orderings
library(utilities)
ORD <- sample.all(5, n = 5)

#Determine compatibility
m <- nrow(ORD)
COMPATIBLE <- rep(FALSE, m)
for (i in 1:m) {
  if (ORD[i, 1] == 1) { COMPATIBLE[i] <- TRUE }
  if ((ORD[i, 1] == 2)&(ORD[i, 2] == 1)) { COMPATIBLE[i] <- TRUE } }

#Compute probability of compatibility
sum(COMPATIBLE)/m

[1] 0.25

