Patrick and Rachel go to a tennis tournament with 7 other couples. Each round is a single's match (1 vs 1). Nobody plays against his/her partner and nobody plays twice against the same player. At the end of the tournament Patrick asked everyone how many people they played against, and found that each one answered with a different number. How many matches did Rachel play?
First notice the wording of the question: "Patrick asked everyone how many people they played against, and found that each one answered with a different number"
This means that
Patrick himself may have played the same number of games as one of the other players.
In particular, as noted in the comments
Nobody could have played 15 matches, hence Patrick heard each answer between 0 and 14 exactly once.
Now observe that
The person who played 14 matches plays everyone except their partner and so their partner must be the one who plays 0 matches (since every other player plays at least one match). Similarly, the person who played 13 matches plays everyone except the person who played 0 and their partner so must be coupled with the person who played 1 match.
Following this line of reasoning further, the person who played 12 matches is coupled with the person who played 2, the person who played 11 is coupled with the person who played 3... and the person who played 8 must be coupled with the person who played 6.
Finally, the person who played 7 matches must be coupled with the remaining person, Patrick, who must also have played 7.
Hence, we deduce that Rachel played
A less conventional solution:
If you subtract the number of played matches from 15 (the maximum), you'll still get the same inequalities but with the paired contestants playing with each other this time. The max number of the total sum of possible matches is $16*15 = 240$, and the new sum exceeds the old by $8*2=16$, so the old sum is $112$, making Patrick a 7-time player. If Rachel has played $r$ times, $15-r=r+1$, meaning $r=7$.
Test if the solution works:
14: Plays with the people from 1 to 6, then both Patrick and Rachel, and from 8 to 15. The only person they don't play with is 0, who's their partner.
13: Plays with the people from 2 to 6, then both Patrick and Rachel, and from 8 to 15. The only remaining person they don't play with is 1, who's their partner.
Same for 12-2 to 8-6. The Patrick - Rachel pair (7-7) also works.
PS. Assume there can be multiple possible solutions:
Let the possible sums be $s_1, s_2,...$ and $s_n$. Then $224-s_1, 224-s_2,...$ and $224-s_n$ are also possible. We can let $s_1+s_2=s_3+s_4=...=224$. So if Patrick can play $p$ matches, he can also play $14-p$, and $p_1+p_2=p_3+p_4=...=14$ (same for Rachel's $r$ matches and 14). If $p =/= r$ both can be anything due to symmetry, which is wrong because $p$ must be odd for $s$ to be even, so $p = r$. A smaller configuration can be achieved if a pair with different numbers of matches is ignored, which means the sum of their number of matches must be 14. $p=r=7$.