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

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How many subsets of  <math>\{2,3,4,5,6,7,8,9\}</math>  contain at least one prime number?
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==Problem==
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<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>
 
  
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==<math>\Large</math> See Also==
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How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?
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{{AMC12 box|year=2015|ab=B|num-a=6|num-b=4}}
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<math>
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\textbf{(A) } 128 \qquad
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\textbf{(B) } 192 \qquad
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\textbf{(C) } 224 \qquad
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\textbf{(D) } 240 \qquad
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\textbf{(E) } 256
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</math>
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==Solution 1 ==
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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.
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==Solution 2==
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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.
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==Solution 3==
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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
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== Video Solution by OmegaLearn ==
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https://youtu.be/8WrdYLw9_ns?t=253
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~ pi_is_3.14
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/3RPbjmksk6w
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~Education, the Study of Everything
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==See Also==
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{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[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
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All AMC 12 Problems and Solutions

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