Difference between revisions of "2018 AIME I Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
| − | + | We 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, so times two. | 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. | ||
| Line 20: | Line 20: | ||
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 all them, since they can be reversed and the 5's can be red or green, we get, after simplifying: <cmath>\boxed{\dfrac{31}{126}}.</cmath> | + | Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: <cmath>\boxed{\dfrac{31}{126}}.</cmath> |
==Solution 2== | ==Solution 2== | ||
Revision as of 10:25, 8 March 2018
Contents
Question
Kathy has 
red cards and green cards. She shuffles the 
cards and lays out 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 
or 
will make Kathy happy, but 
will not. The probability that Kathy will be happy is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
We have 
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 
, since we are choosing a card without replacement.
Then, we have for the numerator for the two of all red and green: ![\[5\cdot 4\cdot 3\cdot 2\cdot 1.\]](http://latex.artofproblemsolving.com/3/4/1/3419852a0b0d3831fa6bb2129a3e743aac605b9a.png)
For the 4 and 1, we have: ![\[5\cdot 4\cdot 3\cdot 2\cdot 5.\]](http://latex.artofproblemsolving.com/c/8/9/c89fc6234333fb4280de8c87026afdfdedc32d42.png)
For the 3 and 2, we have: ![\[5\cdot 4\cdot 3\cdot 5\cdot 4.\]](http://latex.artofproblemsolving.com/d/e/9/de94833d7ba32a429fd6bda47910fa1dd3570f89.png)
For the 2 and 3, we have: ![\[5\cdot 4\cdot 5\cdot 4\cdot 3.\]](http://latex.artofproblemsolving.com/e/3/8/e384f2b6abb4966839e07e8ade21b2eef26e7f53.png)
For the 1 and 4, we have: ![\[5\cdot 5\cdot 4\cdot 3\cdot 2.\]](http://latex.artofproblemsolving.com/9/9/1/991d610c64deeb2defc9ff89e1de8109c7959128.png)
Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: ![\[\boxed{\dfrac{31}{126}}.\]](http://latex.artofproblemsolving.com/0/d/5/0d5ed86fb7ccebf8c013047caf62e7cf31272693.png)
Solution 2
Our probability will be 
First of all, we have choices for the first card, 
choices for the second card, 
choices for the third card, 
choices for the fourth card, and 
choices for the last card. This gives a total of possible ways for five cards to be chosen.
Finding the number of configurations that make Kathy happy is a more difficult task, however, and we will resort to casework to do it.
First, let's look at the appearances of the "happy configurations" that Kathy likes. Based on the premise of the problem, we can realize that there are ten cases for the appearance of the configurations: ![\[\text{RRRRR},\]](http://latex.artofproblemsolving.com/a/6/c/a6cbcfbe2f3476f23b25b6116816621f4117df2b.png)
![\[\text{GGGGG},\]](http://latex.artofproblemsolving.com/3/6/f/36ff8002a0df8335acc374eaec16f99e5ca2458b.png)
![\[\text{RRRRG},\]](http://latex.artofproblemsolving.com/4/1/6/4165bf17e4a524844d648a62ebca0fe9a15fbbd2.png)
![\[\text{GGGGR},\]](http://latex.artofproblemsolving.com/b/c/a/bcaca541803bd93d11a43d36c70dd1fd454db559.png)
![\[\text{RRRGG},\]](http://latex.artofproblemsolving.com/2/a/f/2af1db30e02317880c3e18d9419a7428e7750a5e.png)
![\[\text{GGGRR},\]](http://latex.artofproblemsolving.com/4/9/5/495ccadc22109fa4d988653cca78c58f3e8f931d.png)
![\[\text{RRGGG},\]](http://latex.artofproblemsolving.com/f/7/2/f7212efc150df356a3bd10d96d13b0d93e31c5eb.png)
![\[\text{GGRRR},\]](http://latex.artofproblemsolving.com/2/5/b/25b3716ef083836f9a9aa2ddb5e84590fe90dc2e.png)
![\[\text{RGGGG},\]](http://latex.artofproblemsolving.com/c/c/6/cc6a763966a8bdfc8fb8955d481fbb5cea76b1f2.png)
![\[\text{GRRRR}.\]](http://latex.artofproblemsolving.com/2/3/1/231d6bd562ed25424c63f6f15d6a07d561396d09.png)
But this doesn't mean there are 10 "happy configurations" in total-- remember that we've been treating these cards as distinguishable, so we must continue to do so.
To lighten the load of 10 cases on the human brain, we can note that in the eyes of what we will soon do, and 
are effectively equivalent, and therefore may be treated in the same case. We will have to multiply by 2 at the end, though.
Similarly, we can equate 
and 
as well as 
and 
so that we just have three cases. We can approach each of these cases with constructive counting.
Case 1: -type.
For this case, there are available choices for the first card, 
available choices for the second card, 
for the third card, 
for the fourth card, and 
for the last card. This leads to a total of 
configurations for this case. There are cases of this type.
Case 2: 
-type.
For this case, there are available choices for the first card, available choices for the second card, for the third card, for the fourth card, and choices for the last card (not , because we're doing a new color). This leads to a total of 
configurations for this case. There are cases of this type.
Case 3: 
-type.
For this case, there are available choices for the first card, available choices for the second card, for the third card, for the fourth card, and choices for the last card. This leads to a total of 
configurations for this case. There are cases of this type.
Adding the cases up gives 
"happy" configurations in total.
This means that the probability that Kathy is happy will be 
which simplifies to 
So the answer is 
See Also
| 2018 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
