Probability that at least two of the three had been sitting next to each other. Where am I wrong? 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. What is P?
I calculated by complement method. But I got it wrong.
The total number of ways to select $3$ out of $25$ knights is $C(25,3)$.
Now there are $22$ places to place $A$, followed by $21$ places to place $B$, and $20$ places to place $C$ after $A$ and $B$. Hence, there are ($22\cdot21\cdot20)/3!$ ways to place $A, B, C$ in between these people with restrictions.
If I calculate the probability the answer does not match, what am I getting wrong? 
 A: Your numerator is incorrect.
We count the number of cases in which no two adjacent knights are selected by determining the number of favorable cases for a row, then subtracting those in which two people would be adjacent if the ends of the row are joined.
We arrange $22$ blue balls and $3$ green balls in a row so that no two of the three green balls are consecutive.  Place $22$ blue balls in a row.  This creates $23$ spaces, $21$ between successive balls and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square b \square$$
To ensure that no two of the green balls are consecutive, we choose three of these $23$ spaces in which to place a green ball, which can be done in $\binom{23}{3}$ ways.  
However, if both spaces at the ends of the row are occupied by a green ball, they will be adjacent when the ends of the row are connected to form a circle. Thus, we must subtract these cases.  If both ends of the row are occupied by a green ball and no two green balls are consecutive, then one of the $21$ spaces between successive blue balls must be occupied by a green ball. 
Hence, the number of cases in which no two adjacent knights at the round table are selected is
$$\binom{23}{3} - \binom{21}{1}$$
Dividing by the $\binom{25}{3}$ possible selections of three knights gives the probability
$$\frac{\dbinom{23}{3} - \dbinom{21}{1}}{\dbinom{25}{3}} = \frac{35}{46}$$
that no two consecutive knights are selected, so the probability that at least two adjacent knights are selected is 
$$1 - \frac{35}{46} = \frac{11}{46}$$
We could also count directly.  There are $25$ ways to select three consecutive knights and $25 \cdot 21$ ways to select a pair of adjacent knights (pick the leftmost member of the pair, then select the third knight from among the $21$ knights not adjacent to the pair), for a total of $25 + 25 \cdot 21$ favorable cases, which gives the probability
$$\frac{25 + 25 \cdot 21}{\dbinom{25}{3}} = \frac{11}{46}$$
that at least two consecutive knights are selected.  
