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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 23"

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==Solution 1==
 
==Solution 1==
 
There are <math>29</math> possible pairs of consecutive integers, namely <math>p_1=\{1,2\},  \cdots, p_{29}=\{29,30\}</math>.  
 
There are <math>29</math> possible pairs of consecutive integers, namely <math>p_1=\{1,2\},  \cdots, p_{29}=\{29,30\}</math>.  
−
Let <math>X_i</math> be a random variable with value <math>X_i=1</math> if <math>p_i</math> is part of the 5-element subset, and 0 otherwise.  
+
Define a random variable <math>X_i</math>, with <math>X_i=1</math>, if <math>p_i</math> is part of the 5-element subset, and <math>0</math> otherwise.  
−
The number of pairs of consecutive integers in a 5-element selection is given by the sum <math>X_1+\cdots + X_{29}</math>. By linearity of expectation, the expected value is equal to the sum of the <math>\mathbb{E}[X_i]</math>:
+
Then the number of pairs of consecutive integers in a <math>5</math>-element selection is given by the sum <math>X_1+\cdots + X_{29}</math>. By linearity of expectation, the expected value is equal to the sum of the <math>\mathbb{E}[X_i]</math>:
 
<cmath>\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]</cmath>
 
<cmath>\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]</cmath>
−
To compute <math>\mathbb{E}[X_i]</math>, note that the number of selections for which <math>X_i=1</math> is <math>\binom{28}{3}</math> out of <math>\binom{30}{5}</math> possible selections. Thus<cmath>\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.</cmath>
+
To compute <math>\mathbb{E}[X_i]</math>, note that <math>X_i=1</math> for a total of <math>_{28}C_3</math> out of <math>_{30}C_5</math> possible selections. Thus<cmath>\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.</cmath>
 
~kingofpineapplz
 
~kingofpineapplz
  
Line 18: Line 18:
 
~MT
 
~MT
  
−
== Solution 3 ==
+
==Solution 3 (casework bash)==
 +
 
 +
We will proceed with some casework.
 +
Let <math>n</math> be the number of sets of consecutive numbers in the subset.
 +
Note that the maximum possible value of <math>n</math> is <math>4.</math>
 +
Case <math>1</math>: <math>n = 4</math>
 +
There is only one way to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{1} = 26</math> ways to arrange where that block of <math>5</math> numbers will go, so a total of <math>1 \cdot 26 = 26</math> ways for this case.
 +
Case <math>2</math>: <math>n = 3</math>
 +
There are <math>4</math> ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are <math>\dbinom{26}{2}</math> 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: <math>4 \cdot 325 = 1300.</math>
 +
Case <math>3</math>: <math>n = 2</math>
 +
By Stars and Bars, there are <math>\dbinom{4}{2} = 6</math> ways to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{3}</math> 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: <math>6 \cdot 2600 = 15600.</math>
 +
Case <math>4</math>: <math>n = 1</math>
 +
By Stars and Bars, there are <math>\dbinom{4}{3} = 4</math> ways to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{4}</math> 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: <math>4 \cdot 14950 = 59800.</math>
 +
Since the last case <math>n=0</math> doesn't contribute to the expected value sum, we can ignore it.
 +
Using the expected value formula, our desired value is <cmath>\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.</cmath>
 +
 
 +
-fidgetboss_4000
 +
 
 +
== Solution 4 ==
 
We define an outcome as <math>\left( a_1 ,\cdots, a_5 \right)</math> with <math>1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 30</math>.
 
We define an outcome as <math>\left( a_1 ,\cdots, a_5 \right)</math> with <math>1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 30</math>.
  
Line 363: Line 381:
 
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 4 (casework bash)==
 
  
−
We will proceed with some casework.  
+
~Steven Chen (www.professorchenedu.com)
−
Let <math>n</math> be the number of sets of consecutive numbers in the subset.
+
 
−
Note that the maximum possible value of <math>n</math> is <math>4.</math>
+
==Solution 5==
−
Case <math>1</math>: <math>n = 4</math>
+
Let <math>a_1, a_2, a_3, a_4, a_5</math> be the five numbers chosen. Then, we can consider the first number and the four differences, <math>a_1, a_2-a_1, a_3-a_2, a_4-a_3, a_5-a_4</math>. Now, each must be positive integers and the sum is less than or equal to <math>30</math>, which there are <math>\binom{30}{5}</math> by Hockey-Stick Identity. Now, we can use the Linearity of Expectation on <math>a_2-a_1, a_3-a_2, a_4-a_3, a_5-a_4</math> since each of them being <math>1</math> represents a pair of consecutive integer. For each of them, we have <math>\binom{28}{3}+\binom{27}{3}+...+1=\binom{29}{4}</math>. Now, we have four such numbers to consider, and by Linearity of Expectation, its <math>4\cdot \binom{29}{4}</math>. Now, we divide by <math>\binom{30}{5}</math> to get the answer of <math>\boxed{\textbf{(A)} ~\frac{2}{3}}</math>.
−
There is only one way to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{1} = 26</math> ways to arrange where that block of <math>5</math> numbers will go, so a total of <math>1 \cdot 26 = 26</math> ways for this case.
+
 
−
Case <math>2</math>: <math>n = 3</math>
+
~Hayabusa1
−
There are <math>4</math> ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are <math>\dbinom{26}{2}</math> 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: <math>4 \cdot 325 = 1300.</math>
+
 
−
Case <math>3</math>: <math>n = 2</math>
+
== Video Solution by OmegaLearn ==
−
By Stars and Bars, there are <math>\dbinom{4}{2} = 6</math> ways to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{3}</math> 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: <math>6 \cdot 2600 = 15600.</math>
+
https://youtu.be/EE-TtptBHeI?t=541
−
Case <math>4</math>: <math>n = 1</math>
+
 
−
By Stars and Bars, there are <math>\dbinom{4}{3} = 4</math> ways to arrange the distribution of the number of elements in each block here. There are <math>\dbinom{26}{4}</math> 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: <math>4 \cdot 14950 = 59800.</math>
+
~ pi_is_3.14
−
Since the last case <math>n=0</math> doesn't contribute to the expected value sum, we can ignore it.
+
 
−
Using the expected value formula, our desired value is <cmath>\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.</cmath>
+
==Video Solution==
 +
https://youtu.be/OH5H-ic8Pgw
 +
 
 +
~MathProblemSolvingSkills.com
  
−
-fidgetboss_4000
+
==Video Solution by The Power Of Logic==
  
−
~Steven Chen (www.professorchenedu.com)
+
https://youtu.be/hgsPW2sM1GQ
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-b=22|num-a=24}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:38, 5 January 2023

Problem

What is the average number of pairs of consecutive integers in a randomly selected subset of $5$ distinct integers chosen from the set $\{ 1, 2, 3, …, 30\}$? (For example the set $\{1, 17, 18, 19, 30\}$ has $2$ pairs of consecutive integers.)

$\textbf{(A)}\ \frac{2}{3} \qquad\textbf{(B)}\ \frac{29}{36} \qquad\textbf{(C)}\ \frac{5}{6} \qquad\textbf{(D)}\ \frac{29}{30} \qquad\textbf{(E)}\ 1$

Solution 1

There are $29$ possible pairs of consecutive integers, namely $p_1=\{1,2\}, \cdots, p_{29}=\{29,30\}$. Define a random variable $X_i$, with $X_i=1$, if $p_i$ is part of the 5-element subset, and $0$ otherwise. Then the number of pairs of consecutive integers in a $5$-element selection is given by the sum $X_1+\cdots + X_{29}$. By linearity of expectation, the expected value is equal to the sum of the $\mathbb{E}[X_i]$: \[\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]\] To compute $\mathbb{E}[X_i]$, note that $X_i=1$ for a total of $_{28}C_3$ out of $_{30}C_5$ possible selections. Thus\[\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.\] ~kingofpineapplz

Solution 2

Alternatively, using linearity of expectation on the chosen subset: there are ${5 \choose 2}$ pairs of integers. There are 29 pairs of consecutive integers out of ${30 \choose 2}$ possible pairs, thus the probability that any given pair is consecutive is $\frac{29}{{30 \choose 2}}$. By linearity of expectation, there are $\frac{{5 \choose 2}29}{{30 \choose 2}} = \frac{10 \cdot 29}{\frac{30 \cdot 29}{2}} = \boxed{\textbf{(A)}\ \frac{2}{3}}$ such pairs on average.

~MT

Solution 3 (casework bash)

We will proceed with some casework. Let $n$ be the number of sets of consecutive numbers in the subset. Note that the maximum possible value of $n$ is $4.$ Case $1$: $n = 4$ There is only one way to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{1} = 26$ ways to arrange where that block of $5$ numbers will go, so a total of $1 \cdot 26 = 26$ ways for this case. Case $2$: $n = 3$ There are $4$ ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are $\dbinom{26}{2}$ 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: $4 \cdot 325 = 1300.$ Case $3$: $n = 2$ By Stars and Bars, there are $\dbinom{4}{2} = 6$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{3}$ 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: $6 \cdot 2600 = 15600.$ Case $4$: $n = 1$ By Stars and Bars, there are $\dbinom{4}{3} = 4$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{4}$ 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: $4 \cdot 14950 = 59800.$ Since the last case $n=0$ doesn't contribute to the expected value sum, we can ignore it. Using the expected value formula, our desired value is \[\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.\]

-fidgetboss_4000

Solution 4

We define an outcome as $\left( a_1 ,\cdots, a_5 \right)$ with $1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 30$.

We denote by $\Omega$ the sample space. Hence. $| \Omega | = \binom{30}{5}$.

$\textbf{Case 1}$: There is only 1 pair of consecutive integers.

$\textbf{Case 1.1}$: $\left( a_1 , a_2 \right)$ is the single pair of consecutive integers.

We denote by $E_{11}$ the collection of outcomes satisfying this condition. Hence, $| E_{11} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_3 \geq a_1 + 3 \\ a_4 \geq a_3 + 2 \\ a_5 \geq a_4 + 2 \\ a_5 \leq 30 \\ a_1, a_3, a_4, a_5 \in \Bbb N \end{array} \right.. \]

Denote $b_1 = a_1 - 1$, $b_2 = a_3 - a_1 - 3$, $b_3 = a_4 - a_3 - 2$, $b_4 = a_5 - a_4 - 2$, $b_5 = 30 - a_5$. Hence, $| E_{11} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} b_1 + b_2 + b_3 + b_4 + b_5 = 22 \\ b_1, b_2 , b_3, b_4, b_5 \mbox{ are non-negative integers } \end{array} \right.. \]

Therefore, $| E_{11} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}$.

$\textbf{Case 1.2}$: $\left( a_2 , a_3 \right)$ is the single pair of consecutive integers.

We denote by $E_{12}$ the collection of outcomes satisfying this condition. Hence, $| E_{12} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_4 \geq a_2 + 3 \\ a_5 \geq a_4 + 2 \\ a_5 \leq 30 \\ a_1, a_2, a_4, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{12} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}$.

$\textbf{Case 1.3}$: $\left( a_3 , a_4 \right)$ is the single pair of consecutive integers.

We denote by $E_{13}$ the collection of outcomes satisfying this condition. Hence, $| E_{13} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_3 \geq a_2 + 2 \\ a_5 \geq a_3 + 3 \\ a_5 \leq 30 \\ a_1, a_2, a_3, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{13} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}$.

$\textbf{Case 1.4}$: $\left( a_4 , a_5 \right)$ is the single pair of consecutive integers.

We denote by $E_{14}$ the collection of outcomes satisfying this condition. Hence, $| E_{14} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_3 \geq a_2 + 2 \\ a_4 \geq a_3 + 2 \\ a_4 \leq 29 \\ a_1, a_2, a_3, a_4 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{14} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}$.

$\textbf{Case 2}$: There are 2 pairs of consecutive integers.

$\textbf{Case 2.1}$: $\left( a_1 , a_2 \right)$ and $\left( a_2 , a_3 \right)$ are two pairs of consecutive integers.

We denote by $E_{21}$ the collection of outcomes satisfying this condition. Hence, $| E_{21} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_4 \geq a_1 + 4 \\ a_5 \geq a_4 + 2 \\ a_5 \leq 30 \\ a_1, a_4, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{21} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 2.2}$: $\left( a_1 , a_2 \right)$ and $\left( a_3 , a_4 \right)$ are two pairs of consecutive integers.

We denote by $E_{22}$ the collection of outcomes satisfying this condition. Hence, $| E_{22} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_3 \geq a_1 + 3 \\ a_5 \geq a_3 + 3 \\ a_5 \leq 30 \\ a_1, a_3, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{22} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 2.3}$: $\left( a_1 , a_2 \right)$ and $\left( a_4 , a_5 \right)$ are two pairs of consecutive integers.

We denote by $E_{23}$ the collection of outcomes satisfying this condition. Hence, $| E_{23} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_3 \geq a_1 + 3 \\ a_4 \geq a_3 + 2 \\ a_4 \leq 29 \\ a_1, a_3, a_4 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{23} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 2.4}$: $\left( a_2 , a_3 \right)$ and $\left( a_3 , a_4 \right)$ are two pairs of consecutive integers.

We denote by $E_{24}$ the collection of outcomes satisfying this condition. Hence, $| E_{24} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_5 \geq a_2 + 4 \\ a_5 \leq 30 \\ a_1, a_2, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{24} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 2.5}$: $\left( a_2 , a_3 \right)$ and $\left( a_4 , a_5 \right)$ are two pairs of consecutive integers.

We denote by $E_{25}$ the collection of outcomes satisfying this condition. Hence, $| E_{25} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_4 \geq a_2 + 3 \\ a_4 \leq 29 \\ a_1, a_2, a_4 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{25} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 2.6}$: $\left( a_3 , a_4 \right)$ and $\left( a_4 , a_5 \right)$ are two pairs of consecutive integers.

We denote by $E_{26}$ the collection of outcomes satisfying this condition. Hence, $| E_{26} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_3 \geq a_2 + 2 \\ a_3 \leq 28 \\ a_1, a_2, a_3 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{26} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}$.

$\textbf{Case 3}$: There are 3 pairs of consecutive integers.

$\textbf{Case 3.1}$: $\left( a_1 , a_2 \right)$, $\left( a_2 , a_3 \right)$ and $\left( a_3 , a_4 \right)$ are three pairs of consecutive integers.

We denote by $E_{31}$ the collection of outcomes satisfying this condition. Hence, $| E_{31} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_5 \geq a_1 + 5 \\ a_5 \leq 30 \\ a_1, a_5 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{31} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}$.

$\textbf{Case 3.2}$: $\left( a_1 , a_2 \right)$, $\left( a_2 , a_3 \right)$ and $\left( a_4 , a_5 \right)$ are three pairs of consecutive integers.

We denote by $E_{32}$ the collection of outcomes satisfying this condition. Hence, $| E_{32} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_4 \geq a_1 + 4 \\ a_4 \leq 29 \\ a_1, a_4 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{32} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}$.

$\textbf{Case 3.3}$: $\left( a_1 , a_2 \right)$, $\left( a_3 , a_4 \right)$ and $\left( a_4 , a_5 \right)$ are three pairs of consecutive integers.

We denote by $E_{33}$ the collection of outcomes satisfying this condition. Hence, $| E_{33} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_3 \geq a_1 + 3 \\ a_3 \leq 28 \\ a_1, a_3 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{33} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}$.

$\textbf{Case 3.4}$: $\left( a_2 , a_3 \right)$, $\left( a_3 , a_4 \right)$ and $\left( a_4 , a_5 \right)$ are three pairs of consecutive integers.

We denote by $E_{34}$ the collection of outcomes satisfying this condition. Hence, $| E_{34} |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_2 \leq 27 \\ a_1, a_2 \in \Bbb N \end{array} \right.. \]

Similar to our analysis for Case 1.1, $| E_{34} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}$.

$\textbf{Case 4}$: There are 4 pairs of consecutive integers.

In this case, $\left( a_1, a_2 , a_3 , a_4 , a_5 \right)$ are consecutive integers.

We denote by $E_4$ the collection of outcomes satisfying this condition. Hence, $| E_4 |$ is the number of outcomes satisfying \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_1 \leq 27 \\ a_1 \in \Bbb N \end{array} \right.. \]

Hence, $| E_4 | = 26$.

Therefore, the average number of pairs of consecutive integers is \begin{align*} & \frac{1}{| \Omega|} \left( 1 \cdot \sum_{i=1}^4 | E_{1i} | + 2 \cdot \sum_{i=1}^6 | E_{2i} | + 3 \cdot \sum_{i=1}^4 | E_{3i} | + 4 \cdot | E_4 | \right) \\ & = \frac{1}{\binom{30}{5}} \left( 4 \binom{26}{4} + 12 \binom{26}{3} + 12 \binom{26}{2} + 4 \cdot 26 \right) \\ & = \frac{2}{3} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(A) }\frac{2}{3}}$.


~Steven Chen (www.professorchenedu.com)

Solution 5

Let $a_1, a_2, a_3, a_4, a_5$ be the five numbers chosen. Then, we can consider the first number and the four differences, $a_1, a_2-a_1, a_3-a_2, a_4-a_3, a_5-a_4$. Now, each must be positive integers and the sum is less than or equal to $30$, which there are $\binom{30}{5}$ by Hockey-Stick Identity. Now, we can use the Linearity of Expectation on $a_2-a_1, a_3-a_2, a_4-a_3, a_5-a_4$ since each of them being $1$ represents a pair of consecutive integer. For each of them, we have $\binom{28}{3}+\binom{27}{3}+...+1=\binom{29}{4}$. Now, we have four such numbers to consider, and by Linearity of Expectation, its $4\cdot \binom{29}{4}$. Now, we divide by $\binom{30}{5}$ to get the answer of $\boxed{\textbf{(A)} ~\frac{2}{3}}$.

~Hayabusa1

Video Solution by OmegaLearn

https://youtu.be/EE-TtptBHeI?t=541

~ pi_is_3.14

Video Solution

https://youtu.be/OH5H-ic8Pgw

~MathProblemSolvingSkills.com

Video Solution by The Power Of Logic

https://youtu.be/hgsPW2sM1GQ

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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