Art of Problem Solving

2018 AMC 12B Problems/Problem 5: Difference between revisions

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==Problem==
==Problem==


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


<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>
\textbf{(A) } 128 \qquad
\textbf{(B) } 192 \qquad
\textbf{(C) } 224 \qquad
\textbf{(D) } 240 \qquad
\textbf{(E) } 256
</math>


==Solution 1 ==
==Solution 1 ==
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. 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 <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math>
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==
==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*16</math> ways to choose a subset of the eight numbers, or <math>\boxed { (\textbf{D}) 240 }</math> .
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==
==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 (ProblemsAnswer KeyResources)
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

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