Difference between revisions of "2018 AIME I Problems/Problem 3"
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This means that the probability that Kathy is happy will be <math>\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},</math> which simplifies to <math>\dfrac{31}{126}.</math> | This means that the probability that Kathy is happy will be <math>\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},</math> which simplifies to <math>\dfrac{31}{126}.</math> | ||
− | So the answer is <math>31+126=\boxed{157 | + | So the answer is <math>31+126=\boxed{157}</math> |
==Solution 3== | ==Solution 3== |
Revision as of 20:14, 17 February 2019
Problem
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 RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR 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 .
-gorefeebuddie
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
Solution 3
As the problem states, some examples of valid are ,
, and
. Let's use each of these as more general cases.
Let be the case when there are 2 adjacents of one color, and 3 adjacents of the other color. This yields
combinations (
,
,
, and
). The probability of each of these is equal, equating to
, and since there are
combinations, the probability of this case is
Next case is . Let this be when there are 4 adjacents of one color, and 1 individual color. Once again, this yields
combinations (
,
,
, and
). The probability of each is the same, equating to
, and since there are
combinations, the probability of this case is
The final case is , in which there is just an adjacent block of
colors. There are only
combinations this time, each equating to the probability
, and since there are
combinations, the probability of this case is
.
Thus, the total probability 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.