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

m (→‎Problem 23)
Line 14: Line 14:
  
 
~kingofpineapplz
 
~kingofpineapplz
 +
 +
== Solution 2 ==
 +
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 denote by <math>\Omega</math> the sample space. Hence. <math>| \Omega | = \binom{30}{5}</math>.
 +
 +
<math>\textbf{Case 1}</math>: There is only 1 pair of consecutive integers.
 +
 +
<math>\textbf{Case 1.1}</math>: <math>\left( a_1 , a_2 \right)</math> is the single pair of consecutive integers.
 +
 +
We denote by <math>E_{11}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{11} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_3 \geq a_1 + 3 \\
 +
a_4 \geq a_3 + 2 \\
 +
a_5 \geq a_4 + 2 \\
 +
a_5 \leq 30 \\
 +
a_1, a_3, a_4, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Denote <math>b_1 = a_1 - 1</math>, <math>b_2 = a_3 - a_1 - 3</math>, <math>b_3 = a_4 - a_3 - 2</math>, <math>b_4 = a_5 - a_4 - 2</math>, <math>b_5 = 30 - a_5</math>.
 +
Hence,
 +
<math>| E_{11} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
b_1 + b_2 + b_3 + b_4 + b_5 = 22 \\
 +
b_1, b_2 , b_3, b_4, b_5 \mbox{ are non-negative integers }
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Therefore, <math>| E_{11} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}</math>.
 +
 +
<math>\textbf{Case 1.2}</math>: <math>\left( a_2 , a_3 \right)</math> is the single pair of consecutive integers.
 +
 +
We denote by <math>E_{12}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{12} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_4 \geq a_2 + 3 \\
 +
a_5 \geq a_4 + 2 \\
 +
a_5 \leq 30 \\
 +
a_1, a_2, a_4, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{12} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}</math>.
 +
 +
<math>\textbf{Case 1.3}</math>: <math>\left( a_3 , a_4 \right)</math> is the single pair of consecutive integers.
 +
 +
We denote by <math>E_{13}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{13} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_3 \geq a_2 + 2 \\
 +
a_5 \geq a_3 + 3 \\
 +
a_5 \leq 30 \\
 +
a_1, a_2, a_3, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{13} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}</math>.
 +
 +
<math>\textbf{Case 1.4}</math>: <math>\left( a_4 , a_5 \right)</math> is the single pair of consecutive integers.
 +
 +
We denote by <math>E_{14}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{14} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_3 \geq a_2 + 2 \\
 +
a_4 \geq a_3 + 2 \\
 +
a_4 \leq 29 \\
 +
a_1, a_2, a_3, a_4 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{14} | = \binom{22 + 5 - 1}{5 - 1} = \binom{26}{4}</math>.
 +
 +
<math>\textbf{Case 2}</math>: There are 2 pairs of consecutive integers.
 +
 +
<math>\textbf{Case 2.1}</math>: <math>\left( a_1 , a_2 \right)</math> and <math>\left( a_2 , a_3 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{21}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{21} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_4 \geq a_1 + 4 \\
 +
a_5 \geq a_4 + 2 \\
 +
a_5 \leq 30 \\
 +
a_1, a_4, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{21} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 2.2}</math>: <math>\left( a_1 , a_2 \right)</math> and <math>\left( a_3 , a_4 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{22}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{22} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_3 \geq a_1 + 3 \\
 +
a_5 \geq a_3 + 3 \\
 +
a_5 \leq 30 \\
 +
a_1, a_3, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{22} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 2.3}</math>: <math>\left( a_1 , a_2 \right)</math> and <math>\left( a_4 , a_5 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{23}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{23} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_3 \geq a_1 + 3 \\
 +
a_4 \geq a_3 + 2 \\
 +
a_4 \leq 29 \\
 +
a_1, a_3, a_4 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{23} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 2.4}</math>: <math>\left( a_2 , a_3 \right)</math> and <math>\left( a_3 , a_4 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{24}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{24} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_5 \geq a_2 + 4 \\
 +
a_5 \leq 30 \\
 +
a_1, a_2, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{24} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 2.5}</math>: <math>\left( a_2 , a_3 \right)</math> and <math>\left( a_4 , a_5 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{25}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{25} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_4 \geq a_2 + 3 \\
 +
a_4 \leq 29 \\
 +
a_1, a_2, a_4 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{25} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 2.6}</math>: <math>\left( a_3 , a_4 \right)</math> and <math>\left( a_4 , a_5 \right)</math> are two pairs of consecutive integers.
 +
 +
We denote by <math>E_{26}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{26} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_2 \geq a_1 + 2 \\
 +
a_3 \geq a_2 + 2 \\
 +
a_3 \leq 28 \\
 +
a_1, a_2, a_3 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{26} | = \binom{23 + 4 - 1}{4 - 1} = \binom{26}{3}</math>.
 +
 +
<math>\textbf{Case 3}</math>: There are 3 pairs of consecutive integers.
 +
 +
<math>\textbf{Case 3.1}</math>: <math>\left( a_1 , a_2 \right)</math>, <math>\left( a_2 , a_3 \right)</math> and <math>\left( a_3 , a_4 \right)</math> are three pairs of consecutive integers.
 +
 +
We denote by <math>E_{31}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{31} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_5 \geq a_1 + 5 \\
 +
a_5 \leq 30 \\
 +
a_1, a_5 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{31} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}</math>.
 +
 +
<math>\textbf{Case 3.2}</math>: <math>\left( a_1 , a_2 \right)</math>, <math>\left( a_2 , a_3 \right)</math> and <math>\left( a_4 , a_5 \right)</math> are three pairs of consecutive integers.
 +
 +
We denote by <math>E_{32}</math> the collection of outcomes satisfying this condition.
 +
Hence, <math>| E_{32} |</math> is the number of outcomes satisfying
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
a_1 \geq 1 \\
 +
a_4 \geq a_1 + 4 \\
 +
a_4 \leq 29 \\
 +
a_1, a_4 \in \Bbb N
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Similar to our analysis for Case 1.1,  <math>| E_{32} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}</math>.
 +
 +
4\textbf{Case 3.3}<math>: </math>\left( a_1 , a_2 \right)<math>, </math>\left( a_3 , a_4 \right)<math> and </math>\left( a_4 , a_5 \right)<math> are three pairs of consecutive integers.
 +
 +
We denote by </math>E_{33}<math> the collection of outcomes satisfying this condition.
 +
Hence, </math>| E_{33} |<math> 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,  </math>| E_{33} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}<math>.
 +
 +
</math>\textbf{Case 3.4}<math>: </math>\left( a_2 , a_3 \right)<math>, </math>\left( a_3 , a_4 \right)<math> and </math>\left( a_4 , a_5 \right)<math> are three pairs of consecutive integers.
 +
 +
We denote by </math>E_{34}<math> the collection of outcomes satisfying this condition.
 +
Hence, </math>| E_{34} |<math> 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,  </math>| E_{34} | = \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2}<math>.
 +
 +
</math>\textbf{Case 4}<math>: There are 4 pairs of consecutive integers.
 +
 +
In this case, </math>\left( a_1, a_2 , a_3 , a_4 , a_5 \right)<math> are consecutive integers.
 +
 +
We denote by </math>E_4<math> the collection of outcomes satisfying this condition.
 +
Hence, </math>| E_4 |<math> 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, </math>| E_4 | = 26<math>.
 +
 +
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 </math>\boxed{\textbf{(A) }\frac{2}{3}}$.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:28, 26 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

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)


See Also

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