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Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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<math>\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}</math>
 
<math>\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}</math>
  
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==Solution 1==
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==Solution 1 - Complementary Counting==
  
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We use complementary counting, or Total- what we don't want = what we want. There are a total of <math>2^8</math> ways to create subsets (consider including or excluding each number) and there are a total of <math>2^4</math> subsets only containing composite numbers. Therefore, there are <math>2^8-2^4=240</math> total ways to have at least one prime in a subset.
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We use complementary counting, or  
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<math>\text{total - what we don't want = what we want}</math>.  
 +
 
 +
There are a total of <math>2^8</math> ways to create subsets (consider including or excluding each number) and there are a total of <math>2^4</math> subsets only containing composite numbers (the composite numbers are <math>4, 6, 8, 9</math>). Therefore, there are <math>2^8-2^4=240</math> total ways to have at least one prime in a subset.
  
 
==Solution 2 (Using Answer Choices)==
 
==Solution 2 (Using Answer Choices)==
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-goldenn
 
-goldenn
  
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==Video Solution==
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==Solution 5==
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We multiply the number of possibilities of the set having prime numbers and the set having composites.
 +
 
 +
The possibilities of primes are <math>2^4-1=15</math> (As there is one solution not containing any primes)
 +
 
 +
The possibilities of the set containing composites are <math>2^4=16</math> (There can be a set with no composites)
 +
 
 +
Multiplying this we get <math>15 \cdot 16 = \boxed{\textbf{(D) }240}</math>.
 +
 
 +
-middletonkids
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/RkQPY-_Qieo
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/6Bt4I23JL1s
 +
 
 +
~savannahsolver
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== Video Solution 2==
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https://youtu.be/5UojVH4Cqqs?t=1609
  
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 12:30, 28 May 2023

Problem

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

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

Solution 1 - Complementary Counting

We use complementary counting, or

$\text{total - what we don't want = what we want}$.

There are a total of $2^8$ ways to create subsets (consider including or excluding each number) and there are a total of $2^4$ subsets only containing composite numbers (the composite numbers are $4, 6, 8, 9$). Therefore, there are $2^8-2^4=240$ total ways to have at least one prime in a subset.

Solution 2 (Using Answer Choices)

Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations.

$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$. Using the answer choices, the only multiple of 15 is $\boxed{\textbf{(D) }240}$

By: K6511

Solution 3

Subsets of $\{2,3,4,5,6,7,8,9\}$ include a single digit up to all eight numbers. Therefore, we must add the combinations of all possible subsets and subtract from each of the subsets formed by the composite numbers.

Hence:

$\binom{8}{1} - \binom{4}{1} + \binom{8}{2} - \binom{4}{2} + \binom{8}{3} - \binom{4}{3} + \binom{8}{4} - 1 + \binom{8}{5} + \binom{8}{6} + \binom{8}{7} + 1 = \boxed{\textbf{(D) }240}$

By: pradyrajasai


Solution 4

Total subsets is $(2^8) = 256$ Using complementary counting and finding the sets with composite numbers: only 4,6,8 and 9 are composite. Each one can be either in the set or out: $2^4$ = 16 $256-16=240$ $\boxed{\textbf{(D) }240}$

-goldenn

Solution 5

We multiply the number of possibilities of the set having prime numbers and the set having composites.

The possibilities of primes are $2^4-1=15$ (As there is one solution not containing any primes)

The possibilities of the set containing composites are $2^4=16$ (There can be a set with no composites)

Multiplying this we get $15 \cdot 16 = \boxed{\textbf{(D) }240}$.

-middletonkids

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/RkQPY-_Qieo

~Education, the Study of Everything


Video Solution 1

https://youtu.be/6Bt4I23JL1s

~savannahsolver

Video Solution 2

https://youtu.be/5UojVH4Cqqs?t=1609

~ pi_is_3.14

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
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Problem 6
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