Difference between revisions of "2019 AIME I Problems/Problem 2"

(Solution 3)
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==Solution 2==
 
==Solution 2==
This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</math> ways. Dividing 171 by 380, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=29</math>.
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This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</math> ways. Dividing 171 by 380, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=\boxed{029}</math>.
 
-Fidgetboss_4000
 
-Fidgetboss_4000
  
 
==Solution 3==
 
==Solution 3==
Simply create a grid using graph paper, with 20 columns for J and 20 rows for B. Since we know that <math>B</math> cannot equal <math>C</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=29</math>.
+
Simply create a grid using graph paper, with 20 columns for J and 20 rows for B. Since we know that <math>B</math> cannot equal <math>C</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2019|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:38, 14 March 2019

Problem 2

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Realize that by symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$, which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by 1 (no zero, because the two numbers are distinct). There are $20 * 19 = 380$ total possible combinations of $B$ and $J$, and $1 + 18 * 2 + 1 = 38$ ones that form $P$, so $P = \frac{38}{380} = \frac{1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.

Solution 2

This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$, there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing 171 by 380, our desired probability is $\frac{171}{380} = \frac{9}{20}$. Thus, our answer is $9+20=\boxed{029}$. -Fidgetboss_4000

Solution 3

Simply create a grid using graph paper, with 20 columns for J and 20 rows for B. Since we know that $B$ cannot equal $C$, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$, we can mark the line where $B - J = 2$. Now we sum the number of squares that are on this line and below it. We get $171$. Then we find the number of total squares, which is $400 - 20 = 380$. Finally, we take the ratio $\frac{171}{380}$, which simplifies to $\frac{9}{20}$. Our answer is $9+20=\boxed{029}$.

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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