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

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==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*16</math> ways to choose a subset of the eight numbers, or <math>\boxed { (\textbf{D}) 240 }</math> .
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== Video Solution ==
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https://youtu.be/8WrdYLw9_ns?t=253
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 22:02, 17 January 2021

Problem

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

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

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 $2^8 -2^4 = 240$ subsets with at least 1 prime so the answer is $\Rightarrow \boxed { (\textbf{D}) 240 }\indent$

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 $15*16$ ways to choose a subset of the eight numbers, or $\boxed { (\textbf{D}) 240 }$ .

Video Solution

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

~ pi_is_3.14

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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