Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 12B Problems/Problem 5"

(→‎\Large See Also)
m
 
(23 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
How many subsets of  <math>\{2,3,4,5,6,7,8,9\}</math>  contain at least one prime number?
+
==Problem==
−
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192  \qquad (\text{C}) \indent 224  \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256  </math>
 
  
−
==<math>\Large</math> See Also==
+
How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?
 +
 
 +
<math>
 +
\textbf{(A) } 128 \qquad
 +
\textbf{(B) } 192 \qquad
 +
\textbf{(C) } 224 \qquad
 +
\textbf{(D) } 240 \qquad
 +
\textbf{(E) } 256
 +
</math>
 +
 
 +
==Solution 1 ==
 +
Since an element of a subset is either in or out, the total number of subsets of the <math>8</math>-element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least <math>1</math> prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are <math>4</math> non-primes, there are <math>2^8 -2^4 = \boxed{\textbf{(D) } 240}</math> subsets with at least <math>1</math> prime.
 +
 
 +
==Solution 2==
 +
We can construct our subset by choosing which primes are included and which composites are included. There are <math>2^4-1</math> ways to select the primes (total subsets minus the empty set) and <math>2^4</math> ways to select the composites. Thus, there are <math>15\cdot16=\boxed{\textbf{(D) } 240}</math> ways to choose a subset of the eight numbers.
 +
 
 +
==Solution 3==
 +
We complement count. We know that we have <math>2^8</math> subsets in all, including the empty subset. We subtract <math>\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} = 2^4 = 16</math> We have <math>2^8-16 = 256-16 = \boxed{\textbf{(D) }240}</math> ways to choose a subset of the eight numbers that include a prime number. ~hh99754539
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/8WrdYLw9_ns?t=253
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/3RPbjmksk6w
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==See Also==
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 01:35, 28 May 2023

Problem

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?

$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

Solution 1

Since an element of a subset is either in or out, the total number of subsets of the $8$-element set is $2^8 = 256$. However, since we are only concerned about the subsets with at least $1$ prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are $4$ non-primes, there are $2^8 -2^4 = \boxed{\textbf{(D) } 240}$ subsets with at least $1$ prime.

Solution 2

We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15\cdot16=\boxed{\textbf{(D) } 240}$ ways to choose a subset of the eight numbers.

Solution 3

We complement count. We know that we have $2^8$ subsets in all, including the empty subset. We subtract $\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} = 2^4 = 16$ We have $2^8-16 = 256-16 = \boxed{\textbf{(D) }240}$ ways to choose a subset of the eight numbers that include a prime number. ~hh99754539

Video Solution by OmegaLearn

https://youtu.be/8WrdYLw9_ns?t=253

~ pi_is_3.14

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/3RPbjmksk6w

~Education, the Study of Everything

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&oldid=193609"