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Difference between revisions of "2011 AIME II Problems/Problem 12"
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So, by PIE, the total count of unwanted arrangements is <math>27\cdot 20 - 27\cdot 4 + 18 = 16\cdot27 + 18 = 18\cdot25. </math> So the fraction <cmath> \frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.</cmath> Thus <math>m + n = 56 + 41 = \fbox{097}.</math>
 
So, by PIE, the total count of unwanted arrangements is <math>27\cdot 20 - 27\cdot 4 + 18 = 16\cdot27 + 18 = 18\cdot25. </math> So the fraction <cmath> \frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.</cmath> Thus <math>m + n = 56 + 41 = \fbox{097}.</math>
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==Solution II==
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We should give a specific setup for the table so to simplify. Call countries A, B, C. Someone from A (let me call him/her the "Master A") sits at the northernmost place for configurations. In my solution I choose to represent people by countries, because all the problem says about distinction among delegates is their country. Now, we can see that the total number of ways to arrange is nCr(8, 2) * nCr(6, 3) = 560 (8 seats choose 2 for the rest of the A country, 6 seats choose 3 for B).
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It is not difficult to see that the configurations we don't want are those which put three delegates in a row. Note also that despite my (arbitrary) setting A at the front, cases for A, B, C to be together are symmetric. Now, we use some [[Principle of Inclusion-Exclusion]]. For A three in a row: two cases. First. A, Master A, A: nCr(6, 3) for B seats = <math>20</math>. There's also <math>20</math> for each of the "slant" A sets (i.e. A, A, Master A and Master A, A, A). =<math>60. *3=180</math>.
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For both A and B: Two cases again! A, Master A, A: we see 4 ways for B to have 3-in-a-row. Same for slant = <math>12. *3</math> (AB, BC, or CA) = <math>36</math>.
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For all three: This time, we have 2 for each of the 3 aforementioned A situations = 6. Finally, 180-36+6=150. Those are the ones we don't want. The ones we do want are 410 of 560, so our answer is <math>\fbox{097}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:20, 1 January 2018

Problem 12

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Use complementary probability and Principle of Inclusion-Exclusion. If we consider the delegates from each country to be indistinguishable and number the chairs, we have \[\frac{9!}{(3!)^3} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4}{6\cdot6} = 6\cdot8\cdot7\cdot5 = 30\cdot56\] total ways to seat the candidates.

Of these, there are $3 \times 9 \times \frac{6!}{(3!)^2}$ ways to have the candidates of at least some one country sit together. This comes to \[\frac{27\cdot6\cdot5\cdot4}6 = 27\cdot 20.\]

Among these there are $3 \times 9 \times 4$ ways for candidates from two countries to each sit together. This comes to $27\cdot 4.$

Finally, there are $9 \times 2 = 18.$ ways for the candidates from all the countries to sit in three blocks (9 clockwise arrangements, and 9 counter-clockwise arrangements).

So, by PIE, the total count of unwanted arrangements is $27\cdot 20 - 27\cdot 4 + 18 = 16\cdot27 + 18 = 18\cdot25.$ So the fraction \[\frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.\] Thus $m + n = 56 + 41 = \fbox{097}.$


Solution II

We should give a specific setup for the table so to simplify. Call countries A, B, C. Someone from A (let me call him/her the "Master A") sits at the northernmost place for configurations. In my solution I choose to represent people by countries, because all the problem says about distinction among delegates is their country. Now, we can see that the total number of ways to arrange is nCr(8, 2) * nCr(6, 3) = 560 (8 seats choose 2 for the rest of the A country, 6 seats choose 3 for B).

It is not difficult to see that the configurations we don't want are those which put three delegates in a row. Note also that despite my (arbitrary) setting A at the front, cases for A, B, C to be together are symmetric. Now, we use some Principle of Inclusion-Exclusion. For A three in a row: two cases. First. A, Master A, A: nCr(6, 3) for B seats = $20$. There's also $20$ for each of the "slant" A sets (i.e. A, A, Master A and Master A, A, A). =$60. *3=180$. For both A and B: Two cases again! A, Master A, A: we see 4 ways for B to have 3-in-a-row. Same for slant = $12. *3$ (AB, BC, or CA) = $36$. For all three: This time, we have 2 for each of the 3 aforementioned A situations = 6. Finally, 180-36+6=150. Those are the ones we don't want. The ones we do want are 410 of 560, so our answer is $\fbox{097}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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