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Difference between revisions of "2017 AIME II Problems/Problem 9"

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(→‎Solution)
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==Solution==
 
==Solution==
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<math>\boxed{013}</math>
 
  
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there have to be 2 of 8 card sharing same number and 2 of them sharing same color.
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There have to be <math>2</math> of <math>8</math> cards sharing the same number and <math>2</math> of them sharing same color.
 +
 +
<math>2</math> pairs of cards can't be the same or else there will be <math>2</math> card which are completely same.
  
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and the there 2 pairs of cards can't be all same or there will be 2 card which are completely same
+
WLOG the numbers are <math>1,1,2,3,4,5,6,</math> and <math>7</math> and the colors are <math>a,a,b,c,d,e,f,</math> and <math>g</math>
 +
Then we can get <math>2</math> cases:
  
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WLOG the number are 1,1,2,3,4,5,6,and7 and the color are a,a,b,c,d,e,f,andg
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Case One:
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then we can get 2 cases
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<math>1a,1b,2a,3c,4d,5e,6f,</math> and <math>7g</math>
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in this case, we can discard <math>1a</math>.
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there are <math>2*6=12</math> situations in this case.
  
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1:
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Case Two:
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1a,1b,2a,3c,4d,5e,6f,and 7g
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<math>1b,1c,2a,3a,4d,5e,6f,</math> and <math>7g</math>
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in this case, we can discard 1a.
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In this case, we can't discard.
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there are 2*6=12 situations in this case
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There are <math>\frac{6*5}{2}=15</math> situations in this case
  
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2:
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So the probability is <math>\frac{12}{12+15}=\frac{4}{9}</math>
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1b,1c,2a,3a,4d,5e,6f,and 7g
 
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in this case, we can't discard.
 
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there are (6*5)/2=15 situations in this case
 
  
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so the proprobility is 12/(12+15)=4/9
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The answer is <math>4+9=\boxed{013}</math>
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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 17:24, 23 March 2017

Problem

A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ 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 $\textit{still}$ have at least one card of each color and at least one card with each number if $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

There have to be $2$ of $8$ cards sharing the same number and $2$ of them sharing same color.

$2$ pairs of cards can't be the same or else there will be $2$ card which are completely same.

WLOG the numbers are $1,1,2,3,4,5,6,$ and $7$ and the colors are $a,a,b,c,d,e,f,$ and $g$ Then we can get $2$ cases:

Case One: $1a,1b,2a,3c,4d,5e,6f,$ and $7g$ in this case, we can discard $1a$. there are $2*6=12$ situations in this case.

Case Two: $1b,1c,2a,3a,4d,5e,6f,$ and $7g$ In this case, we can't discard. There are $\frac{6*5}{2}=15$ situations in this case

So the probability is $\frac{12}{12+15}=\frac{4}{9}$

The answer is $4+9=\boxed{013}$

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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