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 $p_1=\{1,2\}, \cdots, p_{29}=\{29,30\}$ . Define a random variable $X_i$ , with $X_i=1$ , if $p_i$ is part of the 5-element subset, and $0$ otherwise. Then the number of pairs of consecutive integers in a $5$ -element selection is given by the sum $X_1+\cdots + X_{29}$ . By linearity of expectation, the expected value is equal to the sum of the $\mathbb{E}[X_i]$ : $\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]$ To compute $\mathbb{E}[X_i]$ , note that $X_i=1$ for a total of $\binom{28}{3}$ out of $\binom{30}{5}$ possible selections. Thus $\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.$ ~kingofpineapplz

Solution 2

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 3 (casework bash)

We will proceed with some casework. Let $n$ be the number of sets of consecutive numbers in the subset. Note that the maximum possible value of $n$ is $4.$ Case $1$ : $n = 4$ There is only one way to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{1} = 26$ ways to arrange where that block of $5$ numbers will go, so a total of $1 \cdot 26 = 26$ ways for this case. Case $2$ : $n = 3$ There are $4$ ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are $\dbinom{26}{2}$ 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: $4 \cdot 325 = 1300.$ Case $3$ : $n = 2$ By Stars and Bars, there are $\dbinom{4}{2} = 6$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{3}$ 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: $6 \cdot 2600 = 15600.$ Case $4$ : $n = 1$ By Stars and Bars, there are $\dbinom{4}{3} = 4$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{4}$ 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: $4 \cdot 14950 = 59800.$ Since the last case $n=0$ doesn't contribute to the expected value sum, we can ignore it. Using the expected value formula, our desired value is $\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.$

-fidgetboss_4000

Solution 4 (casework bash)

We will proceed with some casework. Let $n$ be the number of sets of consecutive numbers in the subset. Note that the maximum possible value of $n$ is $4.$ Case $1$ : $n = 4$ There is only one way to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{1} = 26$ ways to arrange where that block of $5$ numbers will go, so a total of $1 \cdot 26 = 26$ ways for this case. Case $2$ : $n = 3$ There are $4$ ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are $\dbinom{26}{2}$ 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: $4 \cdot 325 = 1300.$ Case $3$ : $n = 2$ By Stars and Bars, there are $\dbinom{4}{2} = 6$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{3}$ 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: $6 \cdot 2600 = 15600.$ Case $4$ : $n = 1$ By Stars and Bars, there are $\dbinom{4}{3} = 4$ ways to arrange the distribution of the number of elements in each block here. There are $\dbinom{26}{4}$ 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: $4 \cdot 14950 = 59800.$ Since the last case $n=0$ doesn't contribute to the expected value sum, we can ignore it. Using the expected value formula, our desired value is $\frac{4 \cdot 26 + 3 \cdot 1300 + 2 \cdot 15600 + 1 \cdot 59800}{\tbinom{30}{5}} = \frac{95004}{142506} = \boxed{\frac{2}{3}}.$

-fidgetboss_4000

~Steven Chen (www.professorchenedu.com)

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
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