Difference between revisions of "2018 AIME I Problems/Problem 3"
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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 all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: < | + | 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: <math>\dfrac{31}{126}</math> |
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+ | Thus the answer is <math>31 + 126 = \boxed{157}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 14:13, 9 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:
For the 4 and 1, we have:
For the 3 and 2, we have:
For the 2 and 3, we have:
For the 1 and 4, we have:
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:
Thus the answer is .
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: 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.