Difference between revisions of "2017 AIME II Problems/Problem 9"
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==Solution== | ==Solution== | ||
<math>\boxed{013}</math> | <math>\boxed{013}</math> | ||
+ | there have to be 2 of 8 card sharing same number and 2 of them sharing same color. | ||
+ | and the there 2 pairs of cards can't be all same or there will be 2 card which are completely same | ||
+ | WLOG the number are 1,1,2,3,4,5,6,and7 and the color are a,a,b,c,d,e,f,andg | ||
+ | then we can get 2 cases | ||
+ | |||
+ | 1: | ||
+ | 1a,1b,2a,3c,4d,5e,6f,and 7g | ||
+ | in this case, we can discard 1a. | ||
+ | there are 2*6=12 situations in this case | ||
+ | |||
+ | 2: | ||
+ | 1b,1c,2a,3a,4d,5e,6f,and 7g | ||
+ | in this case, we can't discard. | ||
+ | there are (6*5)/2=15 situations in this case | ||
+ | |||
+ | so the proprobility is 12/(12+15)=4/9 | ||
+ | |||
+ | the answer is 4+9=013 | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=8|num-a=10}} | {{AIME box|year=2017|n=II|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:45, 23 March 2017
Problem
A special deck of cards contains cards, each labeled with a number from to and colored with one of seven solors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one cardf with each number, the probability that Sharon can discard one of her cards and have at least one card of each color and at least one card with each number if , where and are relatively prime positive integers. Find .
Solution
there have to be 2 of 8 card sharing same number and 2 of them sharing same color. and the there 2 pairs of cards can't be all same or there will be 2 card which are completely same WLOG the number are 1,1,2,3,4,5,6,and7 and the color are a,a,b,c,d,e,f,andg then we can get 2 cases
1: 1a,1b,2a,3c,4d,5e,6f,and 7g in this case, we can discard 1a. there are 2*6=12 situations in this case
2: 1b,1c,2a,3a,4d,5e,6f,and 7g in this case, we can't discard. there are (6*5)/2=15 situations in this case
so the proprobility is 12/(12+15)=4/9
the answer is 4+9=013
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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