Difference between revisions of "2018 AIME I Problems/Problem 3"
(→Solution) |
(→Solution) |
||
Line 23: | Line 23: | ||
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: \dfrac{31}{126}. | + | Adding up and remembering to double the last four cases, since they can be reversed, we get, after simplifying: <math>\dfrac{31}{126}.</math> |
To finish, we add <math>31+126=\boxed{155}</math>. | To finish, we add <math>31+126=\boxed{155}</math>. |
Revision as of 15:10, 7 March 2018
Question
Kathy has
Solution
Light work
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:
To finish, we add .