Difference between revisions of "2019 AIME I Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | This problem is asks 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>. | + | This problem is asks how many ways there are to choose <math>2</math> distinct elements from a <math>20</math> element set such that no <math>2</math> 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 <math>171</math> by <math>380</math>, 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 | ||
Revision as of 12:48, 26 December 2019
Problem 2
Jenn randomly chooses a number from
. Bela then randomly chooses a number
from
distinct from
. The value of
is at least
with a probability that can be expressed in the form
where
and
are relatively prime positive integers. Find
.
Solution
By symmetry, the desired probability is equal to the probability that is at most
, which is
where
is the probability that
and
differ by
(no zero, because the two numbers are distinct). There are
total possible combinations of
and
, and
ones that form
, so
. Therefore the answer is
.
Solution 2
This problem is asks how many ways there are to choose distinct elements from a
element set such that no
elements are adjacent. Using the well-known formula
, there are
ways. Dividing
by
, our desired probability is
. Thus, our answer is
.
-Fidgetboss_4000
Solution 3
Create a grid using graph paper, with 20 columns for the values of J from 1 to 20 and 20 rows for the values of B from 1 to 20. Since cannot equal
, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since
must be at least
, we can mark the line where
. Now we sum the number of squares that are on this line and below it. We get
. Then we find the number of total squares, which is
. Finally, we take the ratio
, which simplifies to
. Our answer is
.
Solution 4
We can see that if B chooses 20, J has options 1-18, such that . If B chooses 19, J has choices 1-17. By continuing this pattern, B will choose 3 and J will have 1 option. Summing up the total, we get
as the total number of solutions. The total amount of choices is
(B and J must choose different numbers), so the probability is
. Therefore, the answer is
-eric2020
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.