Difference between revisions of "2018 AIME I Problems/Problem 3"

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Light work
 
Light work
  
[hide=Solution]
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You have <math>2+4\cdot 2</math> cases total.
 
You have <math>2+4\cdot 2</math> cases total.
  
The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order,
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The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.
so times two.
 
  
 
Obviously the denominator is <math>10\cdot 9\cdot 8\cdot 7\cdot 6</math>, since we are choosing a card without replacement.
 
Obviously the denominator is <math>10\cdot 9\cdot 8\cdot 7\cdot 6</math>, since we are choosing a card without replacement.
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For the 1 and 4, we have: <cmath>5\cdot 5\cdot 4\cdot 3\cdot 2.</cmath>
 
For the 1 and 4, we have: <cmath>5\cdot 5\cdot 4\cdot 3\cdot 2.</cmath>
  
Adding up and remembering to double the last four cases, since they can be reversed, we get, after simplifying: <cmath>\boxed{\dfrac{31}{126}}.</cmath> [/hide]
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Adding up and remembering to double the last four cases, since they can be reversed, we get, after simplifying: <cmath>\boxed{\dfrac{31}{126}}.</cmath>

Revision as of 15:06, 7 March 2018

Question

Kathy has 5 red cards and 5 green cards. She shuffles the 10 cards and lays out 5 of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG,GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is mn, where m and n are relatively prime positive integers. Find m+n.


Solution

Light work


You have $2+4\cdot 2$ cases total.

The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.

Obviously the denominator is $10\cdot 9\cdot 8\cdot 7\cdot 6$, since we are choosing a card without replacement.

Then, we have for the numerator for the two of all red and green: \[5\cdot 4\cdot 3\cdot 2\cdot 1.\]

For the 4 and 1, we have: \[5\cdot 4\cdot 3\cdot 2\cdot 5.\]

For the 3 and 2, we have: \[5\cdot 4\cdot 3\cdot 5\cdot 4.\]

For the 2 and 3, we have: \[5\cdot 4\cdot 5\cdot 4\cdot 3.\]

For the 1 and 4, we have: \[5\cdot 5\cdot 4\cdot 3\cdot 2.\]

Adding up and remembering to double the last four cases, since they can be reversed, we get, after simplifying: \[\boxed{\dfrac{31}{126}}.\]