# 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?

• Can you please elaborate a bit in your third paragraph (line by line)? Then it becomes easier to see what went wrong where. – Imago Jan 19 '19 at 12:32
• @Imago Imagine placing $22$ chairs around the table. To ensure that no two of the three selected people are adjacent, Rituraj chose them by placing the three selected people in three of the $22$ gaps to the immediate right of those chairs (those chairs represent the positions where the knights who are not selected for the mission sit), giving an answer of $$1 - \frac{\binom{22}{3}}{\binom{25}{3}}$$ – N. F. Taussig Jan 19 '19 at 12:35
• I found the solution here artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/… )but i found the total restricted counts by method of solution 1 and total unrestricted counts by method of soution 3. The final answer should be same but it does not come so – Rituraj Tripathy Jan 19 '19 at 12:39
• @N.F.Taussig That needs to be $1-\frac{\frac{25}{22}\binom{22}{3}}{\binom{25}{3}}$ See this recent question as well as its linked and related questions for obtaining the formula $\frac{n}{n-r}\binom{n-r}{r}$ for counting the number of ways in which none are adjacent. – Daniel Mathias Jan 19 '19 at 16:25

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.