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| | ~MT | | ~MT |
| | | | |
| − | == Solution 3 == | + | ==Solution 3 (casework bash)== |
| − | 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>. | + | 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> |
| | | | |
| − | <math>\textbf{Case 1}</math>: There is only 1 pair of consecutive integers.
| + | -fidgetboss_4000 |
| − | | |
| − | <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>.
| |
| − | | |
| − | <math>\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}}</math>.
| |
| | | | |
| | ==Solution 4 (casework bash)== | | ==Solution 4 (casework bash)== |
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 
.
Define a random variable 
, with 
, if 
is part of the 5-element subset, and 
otherwise.
Then the number of pairs of consecutive integers in a -element selection is given by the sum 
. By linearity of expectation, the expected value is equal to the sum of the ![$\mathbb{E}[X_i]$](//latex.artofproblemsolving.com/4/a/5/4a5abedf8107b2fa9f5714fba2748f1ebeff4a3a.png)
:
![\[\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]\]](//latex.artofproblemsolving.com/d/5/6/d562a9c549eb4cdb0efd75d67d9579654ccd272e.png)
To compute , note that for a total of 
out of 
possible selections. Thus![\[\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.\]](//latex.artofproblemsolving.com/0/9/3/09357202a7c095d8f504a6b8bf802da0bcc70f14.png)
~kingofpineapplz
Solution 2
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 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 ![\[\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.\]](//latex.artofproblemsolving.com/4/4/9/449099122d1aefe5cb18f5695067b8b3e014039b.png)
-fidgetboss_4000
Solution 4 (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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
distinct integers chosen from the set 
? (For example the set 
has 
pairs of consecutive integers.)
possible pairs of consecutive integers, namely 
.
Define a random variable 
, with 
, if 
is part of the 5-element subset, and 
otherwise.
Then the number of pairs of consecutive integers in a 
. By linearity of expectation, the expected value is equal to the sum of the ![$\mathbb{E}[X_i]$](http://latex.artofproblemsolving.com/4/a/5/4a5abedf8107b2fa9f5714fba2748f1ebeff4a3a.png)
:
![\[\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]\]](http://latex.artofproblemsolving.com/d/5/6/d562a9c549eb4cdb0efd75d67d9579654ccd272e.png)
To compute 
out of 
possible selections. Thus![\[\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.\]](http://latex.artofproblemsolving.com/0/9/3/09357202a7c095d8f504a6b8bf802da0bcc70f14.png)
~kingofpineapplz
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.
be the number of sets of consecutive numbers in the subset.
Note that the maximum possible value of 
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 
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 ![\[\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.\]](http://latex.artofproblemsolving.com/4/4/9/449099122d1aefe5cb18f5695067b8b3e014039b.png)
