Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 2: edited for clarity)
Line 8: Line 8:
 
==Solution 2==
 
==Solution 2==
  
−
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.
+
First note that out of the <math>8</math> selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these <math>2</math> pairs of cards can't be the same or else there will be <math>2</math> cards which are completely same. Then, WLOG let the numbers be <math>1,1,2,3,4,5,6,</math> and <math>7</math> and the colors be <math>a,a,b,c,d,e,f,</math> and <math>g</math>. We therefore obtain only <math>2</math> cases:
−
 
−
<math>2</math> pairs of cards can't be the same or else there will be <math>2</math> 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>
+
Case One: <math>1a,1b,2a,3c,4d,5e,6f,</math> and <math>7g</math>
−
Then we can get <math>2</math> cases:
+
In this case, we can discard <math>1a</math>.
 +
There are <math>2*6=12</math> situations in this case.
  
−
Case One:
+
Case Two: <math>1b,1c,2a,3a,4d,5e,6f,</math> and <math>7g</math>
−
<math>1a,1b,2a,3c,4d,5e,6f,</math> and <math>7g</math>
 
−
in this case, we can discard <math>1a</math>.
 
−
there are <math>2*6=12</math> situations in this case.
 
−
 
 
−
Case Two:
 
−
<math>1b,1c,2a,3a,4d,5e,6f,</math> and <math>7g</math>
 
 
In this case, we can't discard.
 
In this case, we can't discard.
−
There are <math>\dbinom{6}{2}=15</math> situations in this case
+
There are <math>\dbinom{6}{2}=15</math> situations in this case.
−
 
 
−
So the probability is <math>\frac{12}{12+15}=\frac{4}{9}</math>
 
  
−
The answer is <math>4+9=\boxed{013}</math>
+
So the probability is <math>\frac{12}{12+15}=\frac{4}{9}</math>, giving us the answer of <math>4+9=\boxed{013}</math>.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 01:45, 21 July 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 colors. 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 card 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 is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1

Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$, $1$, $2$, $3$, $4$, $5$, $6$, and $7$, in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$'s. In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$s. The number of ways for this not to happen is ${6\choose2}$, so the number of ways for it to happen is $\left({8\choose2}-1\right)-{6\choose2}$. Each of these assignments is equally likely, so the probability that Sharon can discard one of her cards and still have at least one card of each color and at least one card with each number is $\frac{\left({8\choose2}-1\right)-{6\choose2}}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}$.

Solution 2

First note that out of the $8$ selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these $2$ pairs of cards can't be the same or else there will be $2$ cards which are completely same. Then, WLOG let the numbers be $1,1,2,3,4,5,6,$ and $7$ and the colors be $a,a,b,c,d,e,f,$ and $g$. We therefore obtain only $2$ 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 $\dbinom{6}{2}=15$ situations in this case.

So the probability is $\frac{12}{12+15}=\frac{4}{9}$, giving us the answer of $4+9=\boxed{013}$.

Solution 3

There are $7!$ ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then $42$ cards remaining. Thus, there are $7!(42)$ desired sets.

Now, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of $\binom{7}{2}$ ways to pick 2 numbers, $7$ colors to assign them to, $\binom{6}{2}$ ways to pick 2 nonchosen colors, $5$ ways to pick a number to assign them to, and $4!$ ways to assign the rest.

Thus, the answer is $\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}$. Dividing out $5!$ yields $\frac{42(42)}{42(42) + 21(15)(7)}$ which is equal to $\frac{2(42)}{2(42) + 15(7)}$ which is equal to $\frac{12}{12 + 15}$ which is equal to $\frac{4}{9}$ giving a final answer of $\boxed{013}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_9&oldid=86484"