# 2018 AMC 10B Problems/Problem 5

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

## 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

~savannahsolver

## Video Solution 2

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 