Difference between revisions of "2018 AIME I Problems/Problem 3"
m (Removed protection from "2018 AIME I Problems/Problem 3") |
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+ | ==Question== | ||
+ | Kathy has | ||
+ | |||
+ | ==Solution== | ||
+ | Light work | ||
+ | |||
+ | [hide=Solution] | ||
+ | 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, | ||
+ | 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. | ||
+ | |||
+ | Then, we have for the numerator for the two of all red and green: <cmath>5\cdot 4\cdot 3\cdot 2\cdot 1.</cmath> | ||
+ | |||
+ | For the 4 and 1, we have: <cmath>5\cdot 4\cdot 3\cdot 2\cdot 5.</cmath> | ||
+ | |||
+ | For the 3 and 2, we have: <cmath>5\cdot 4\cdot 3\cdot 5\cdot 4.</cmath> | ||
+ | |||
+ | For the 2 and 3, we have: <cmath>5\cdot 4\cdot 5\cdot 4\cdot 3.</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] |
Revision as of 15:06, 7 March 2018
Question
Kathy has
Solution
Light work
[hide=Solution] You 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 the last four cases, since they can be reversed, we get, after simplifying: [/hide]