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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?

N. F. Taussig
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Rituraj Tripathy
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  • 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
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    @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 https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7 )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
  • Yes but i get the answer wrong. Please help me out. – Rituraj Tripathy Jan 19 '19 at 12:42
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    @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
  • @DanielMathias I suggest that you answer the question. – N. F. Taussig Jan 19 '19 at 16:58

1 Answers1

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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.

N. F. Taussig
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