Difference between revisions of "2018 AMC 12B Problems/Problem 5"
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==Solution 3== | ==Solution 3== | ||
− | We complement count. We know that we have <math>2^8</math> subsets in all, including the empty subset. We | + | 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 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/3RPbjmksk6w | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== |
Latest revision as of 02:35, 28 May 2023
Contents
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
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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 |
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