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

(Problem)
(See Also)
(2 intermediate revisions by one other user not shown)
Line 6: Line 6:
  
 
==Solution 1 ==
 
==Solution 1 ==
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. 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 <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math> (Giraffefun)
+
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. 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 <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math>
  
 
==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> (mathislife16).
+
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> .
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Revision as of 17:24, 18 June 2018

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 $\boxed { (\textbf{D}) 240 }$ .

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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. AMC logo.png