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

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

Revision as of 22:40, 28 November 2021

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

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

$\textbf{Case 3.3}$: $\left( a_1 , a_2 \right)$, $\left( a_3 , a_4 \right)$ 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)

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

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