Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$.
Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place.
We will place $A$, $B$, and $C$ with and without the restriction.
There are $22$ places to put $A$, followed by $21$ places to put $B$, and $20$ places to put $C$ after $A$ and $B$. Hence, there are $22\cdot21\cdot20$ ways to place $A, B, C$ in between these people with restrictions.
Without restrictions, there are $22$ places to put $A$, followed by $23$ places to put $B$, and $24$ places to put $C$ after $A$ and $B$. Hence, there are $22\cdot23\cdot24$ ways to place $A,B,C$ in between these people without restrictions.
Thus, the desired probability is $1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$, and the answer is $11+46=\boxed{57}$.