2021 Fall AMC 12B Problems/Problem 23

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 $\{1,2\}, \{2,3\},\cdots,\{29,30\}$ .

The probability that a certain pair of consecutive integers are in the $5$ integer subset is $\frac5{30}$ for the first number being chosen, multiplied by $\frac4{29}$ 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 $\frac5{30}\cdot\frac4{29}\cdot29=\boxed{\textbf{(A)}\ \frac{2}{3}}.$

~kingofpineapplz

Solution 2

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, $| E_{32} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}$ .

4\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$ (Error compiling LaTeX. Unknown error_msg)E_{33} $the collection of outcomes satisfying this condition. Hence,$ | E_{33} |$is the number of outcomes satisfying <cmath> \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_3 \geq a_1 + 3 \\ a_3 \leq 28 \\ a_1, a_3 \in \Bbb N \end{array} \right.. \] </cmath>

Similar to our analysis for Case 1.1,$ (Error compiling LaTeX. Unknown error_msg)| 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$ (Error compiling LaTeX. Unknown error_msg)E_{34} $the collection of outcomes satisfying this condition. Hence,$ | E_{34} |$is the number of outcomes satisfying <cmath> \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_2 \geq a_1 + 2 \\ a_2 \leq 27 \\ a_1, a_2 \in \Bbb N \end{array} \right.. \] </cmath>

Similar to our analysis for Case 1.1,$ (Error compiling LaTeX. Unknown error_msg)| E_{34} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2} $.$ \textbf{Case 4}$: There are 4 pairs of consecutive integers.

In this case,$ (Error compiling LaTeX. Unknown error_msg)\left( a_1, a_2 , a_3 , a_4 , a_5 \right)$are consecutive integers.

We denote by$ (Error compiling LaTeX. Unknown error_msg)E_4 $the collection of outcomes satisfying this condition. Hence,$ | E_4 |$is the number of outcomes satisfying <cmath> \[ \left\{ \begin{array}{l} a_1 \geq 1 \\ a_1 \leq 27 \\ a_1 \in \Bbb N \end{array} \right.. \] </cmath>

Hence,$ (Error compiling LaTeX. Unknown error_msg)| E_4 | = 26$.

Therefore, the average number of pairs of consecutive integers is <cmath> \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*} </cmath>

Therefore, the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(A) }\frac{2}{3}}$.

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2021 Fall AMC 12B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

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