Difference between revisions of "2021 Fall AMC 12B Problems/Problem 23"
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Therefore, the answer is <math>\boxed{\textbf{(A) }\frac{2}{3}}</math>. | Therefore, the answer is <math>\boxed{\textbf{(A) }\frac{2}{3}}</math>. | ||
− | ==Solution | + | ==Solution 3 (casework bash)== |
We will proceed with some casework. | We will proceed with some casework. |
Revision as of 13:56, 9 December 2021
Problem
What is the average number of pairs of consecutive integers in a randomly selected subset of distinct integers chosen from the set
? (For example the set
has
pairs of consecutive integers.)
Solution 1
There are possible pairs of consecutive integers, namely
.
The probability that a certain pair of consecutive integers are in the integer subset is
for the first number being chosen, multiplied by
for the second number being chosen.
Therefore, by linearity of expectation, the expected number of pairs of consecutive integers in the 5-integer subset is
~kingofpineapplz
Alternatively, using linearity of expectation on the chosen subset: there are pairs of integers. There are 29 pairs of consecutive integers out of
possible pairs, thus the probability that any given pair is consecutive is
. By linearity of expectation, there are
such pairs on average.
~MT
Solution 2
We define an outcome as with
.
We denote by the sample space. Hence.
.
: There is only 1 pair of consecutive integers.
:
is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Denote ,
,
,
,
.
Hence,
is the number of outcomes satisfying
Therefore, .
:
is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 2 pairs of consecutive integers.
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
and
are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 3 pairs of consecutive integers.
:
,
and
are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
,
and
are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
,
and
are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
:
,
and
are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 4 pairs of consecutive integers.
In this case, are consecutive integers.
We denote by the collection of outcomes satisfying this condition.
Hence,
is the number of outcomes satisfying
Hence, .
Therefore, the average number of pairs of consecutive integers is
Therefore, the answer is .
Solution 3 (casework bash)
We will proceed with some casework.
Let be the number of sets of consecutive numbers in the subset.
Note that the maximum possible value of
is
Case
:
There is only one way to arrange the distribution of the number of elements in each block here. There are
ways to arrange where that block of
numbers will go, so a total of
ways for this case.
Case
:
There are
ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are
ways to arrange where those two blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case:
Case
:
By Stars and Bars, there are
ways to arrange the distribution of the number of elements in each block here. There are
ways to arrange where those three blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case:
Case
:
By Stars and Bars, there are
ways to arrange the distribution of the number of elements in each block here. There are
ways to arrange where those four blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case:
Since the last case
doesn't contribute to the expected value sum, we can ignore it.
Using the expected value formula, our desired value is
-fidgetboss_4000
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.