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Difference between revisions of "1983 AIME Problems/Problem 7"

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[Right answer, but ...  Please click discussion tab.]
 
[Right answer, but ...  Please click discussion tab.]
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==Solution 2 (possibly simpler)==
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We will find the probability they are none are sitting next to each other, and subtract 1.
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Imagine the 22 other people are already seated, and fixed into place. (22 indistinguishable people).
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We will place A, B, and C with and without restriction.
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There are <math>22</math> places to place A, followed by <math>21</math> places to place B, and <math>20</math> places to place C after <math>A</math> and <math>B</math>.
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Hence, there are <math>22\cdot21\cdot20</math> ways to place A, B, C in between these people with restrictions.
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Without restrictions, there are <math>22</math> places to palce A, followed by <math>23</math> places to place B, and <math>24</math> places to place C after <math>A</math> and <math>B</math>.
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Hence, there are <math>22\cdot23\cdot24</math> ways to place A,B,C in between these people without restrictions.
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Hence, the desired amount is <math>1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}</math>
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So the answer is <math>11+46=57</math>
  
 
== See also ==
 
== See also ==

Revision as of 22:51, 16 November 2008

Problem

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 of 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 the denominator?

Solution

We can also find the probability that none of them are sitting together, and subtract from 1. There are 2 cases:

  1. They are all separated by at least 2 seats.
  2. One of them is separated from one of the others by 1 seat.

In case (1), the probability is $\frac{25}{25} \cdot \frac{20}{24} \cdot \frac{19}{23} =\frac{380}{552}$.

But what if they're only separated by 1 seat? In that case, the probability is $\frac{25}{25} \cdot \frac{2}{24} \cdot \frac{20}{23}=\frac{40}{552}$.

The total probability is $\frac{420}{552}=\frac{35}{46}$, reducing the fraction gives: $1-\frac{35}{46}=\frac{11}{46}$. The sum of the numerator and denominator is $11+46=57$

[Right answer, but ... Please click discussion tab.]

Solution 2 (possibly simpler)

We will find the probability they are none are sitting next to each other, and subtract 1.

Imagine the 22 other people are already seated, and fixed into place. (22 indistinguishable people).

We will place A, B, and C with and without restriction.

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 $22\cdot21\cdot20$ ways to place A, B, C in between these people with restrictions.

Without restrictions, there are $22$ places to palce A, followed by $23$ places to place B, and $24$ places to place C after $A$ and $B$.

Hence, there are $22\cdot23\cdot24$ ways to place A,B,C in between these people without restrictions.

Hence, the desired amount is $1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$

So the answer is $11+46=57$

See also

1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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