Difference between revisions of "2018 AMC 12B Problems/Problem 5"
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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 | 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 == | + | == Video Solution by OmegaLearn == |
https://youtu.be/8WrdYLw9_ns?t=253 | https://youtu.be/8WrdYLw9_ns?t=253 | ||
Revision as of 01:56, 16 January 2023
Problem
How many subsets of 
contain at least one prime number?
Solution 1
Since an element of a subset is either in or out, the total number of subsets of the 
-element set is 
. However, since we are only concerned about the subsets with at least 
prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 
non-primes, there are 
subsets with at least prime.
Solution 2
We can construct our subset by choosing which primes are included and which composites are included. There are 
ways to select the primes (total subsets minus the empty set) and 
ways to select the composites. Thus, there are 
ways to choose a subset of the eight numbers.
Solution 3
We complement count. We know that we have 
subsets in all, including the empty subset. We subtract 
We have 
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
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. 
