Difference between revisions of "2021 Fall AMC 12B Problems/Problem 23"
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− | ==Problem | + | ==Problem== |
What is the average number of pairs of consecutive integers in a randomly selected subset of <math>5</math> distinct integers chosen from the set <math>\{ 1, 2, 3, …, 30\}</math>? (For example the set <math>\{1, 17, 18, 19, 30\}</math> has <math>2</math> pairs of consecutive integers.) | What is the average number of pairs of consecutive integers in a randomly selected subset of <math>5</math> distinct integers chosen from the set <math>\{ 1, 2, 3, …, 30\}</math>? (For example the set <math>\{1, 17, 18, 19, 30\}</math> has <math>2</math> pairs of consecutive integers.) | ||
Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
− | There are <math>29</math> possible pairs of consecutive integers, namely <math>\{1,2\}, \{2,3\},\cdots,\{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>. |
+ | 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. | ||
+ | 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> | ||
+ | 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 | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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 | ||
+ | |||
+ | ==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 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>. | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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>. | ||
+ | |||
+ | ~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== | ==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
Contents
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 : To compute , note that for a total of out of possible selections. Thus ~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
-fidgetboss_4000
Solution 4
We define an outcome as with .
We denote by the sample space. Hence. .
: There is only 1 pair of consecutive integers.
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Denote , , , , . Hence, is the number of outcomes satisfying
Therefore, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 2 pairs of consecutive integers.
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 3 pairs of consecutive integers.
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 4 pairs of consecutive integers.
In this case, are consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Hence, .
Therefore, the average number of pairs of consecutive integers is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 5
Let be the five numbers chosen. Then, we can consider the first number and the four differences, . Now, each must be positive integers and the sum is less than or equal to , which there are by Hockey-Stick Identity. Now, we can use the Linearity of Expectation on since each of them being represents a pair of consecutive integer. For each of them, we have . Now, we have four such numbers to consider, and by Linearity of Expectation, its . Now, we divide by to get the answer of .
~Hayabusa1
Video Solution by OmegaLearn
https://youtu.be/EE-TtptBHeI?t=541
~ pi_is_3.14
Video Solution
~MathProblemSolvingSkills.com
Video Solution by The Power Of Logic
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
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