Revision as of 15:56, 31 May 2018

Problem

Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ .

Solution 1

The number of subsets of a set with $n$ elements is $2^n$ . The total number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ is equal to $2^8$ . The number of sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ can be found using complementary counting. There are $2^5$ subsets of $\{1, 2, 3, 4, 5\}$ and $2^5$ subsets of $\{4, 5, 6, 7, 8\}$ . It is easy to make the mistake of assuming there are $2^5+2^5$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , but the $2^2$ subsets of $\{4, 5\}$ are overcounted. There are $2^5+2^5-2^2$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , so there are $2^8-(2^5+2^5-2^2)$ subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ . $2^8-(2^5+2^5-2^2)=\boxed{196}$ .

Solution 2

Upon inspection, a viable set must contain at least one element from both of the sets $\{1, 2, 3, 4, 5\}$ and $\{4, 5, 6, 7, 8\}$ . Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is $2^3=8$ , but we want to exclude the empty set, giving us 7 ways to choose from $\{1, 2, 3\}$ or $\{6, 7, 8\}$ . We can take each of these $7 \times 7=49$ sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is $49 \times 4=\boxed{196}$ .

Solution 3

This solution is very similar to Solution $2$ . The set of all subsets of $\{1,2,3,4,5,6,7,8\}$ that are disjoint with respect to $\{4,5\}$ and are not disjoint with respect to the complements of sets (and therefore not a subset of) $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ will be named $S$ , which has $7\cdot7=49$ members. The union of each member in $S$ and the $2^2=4$ subsets of $\{4,5\}$ will be the members of set $Z$ , which has $49\cdot4=\boxed{196}$ members. $\blacksquare$

Solution by a1b2

Solution 4

Consider that we are trying to figure out how many subsets are possible of $\{1,2,3,4,5,6,7,8\}$ that are not in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Assume that the number of numbers we pick from the subset $\{1,2,3,4,5,6,7,8\}$ is $n$ . Thus, we can compute this problem with simple combinatorics:

If $n=1$ , $\binom{8}{1}$ - ( $\binom{5}{1}$ + $\binom{5}{1}$ - 2) [subtract 2 to eliminate the overcounting of the subset $\{4\}$ or $\{5\}$ ] = 8 - 8 = $0$

If $n=2$ , $\binom{8}{2}$ - ( $\binom{5}{2}$ + $\binom{5}{2}$ - 1) [subtract 1 to eliminate the overcounting of the subset $\{4,5\}$ ] = 28 - 19 = $9$

If $n=3$ , $\binom{8}{3}$ - ( $\binom{5}{3}$ + $\binom{5}{3}$ ) = 56 - 20 = $36$

If $n=4$ , $\binom{8}{4}$ - ( $\binom{5}{4}$ + $\binom{5}{4}$ ) = 70 - 10 = $60$

If $n=5$ , $\binom{8}{5}$ - ( $\binom{5}{5}$ + $\binom{5}{5}$ ) = 56 - 2 = $54$

If $n>5$ , then the set $\{1,2,3,4,5,6,7,8\}$ is never in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Thus,

If $n=6$ , $\binom{8}{6}$ = $28$

If $n=7$ , $\binom{8}{7}$ = $8$

If $n=8$ , $\binom{8}{8}$ = $1$

Adding these together, our solution becomes $0$ + $9$ + $36$ + $60$ + $54$ + $28$ + $8$ + $1$ $=\boxed{196}$

Solution by IronicNinja~

@@ Line 13: / Line 13: @@
 Solution by [[User:a1b2|a1b2]]
+==Solution 4==
+Consider that we are trying to figure out how many subsets are possible of <math>\{1,2,3,4,5,6,7,8\}</math> that are not in violation of the two subsets <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math>. Assume that the number of numbers we pick from the subset <math>\{1,2,3,4,5,6,7,8\}</math> is <math>n</math>. Thus, we can compute this problem with simple combinatorics:
+If <math>n=1</math>, <math>\binom{8}{1}</math> - (<math>\binom{5}{1}</math> + <math>\binom{5}{1}</math> - 2) [subtract 2 to eliminate the overcounting of the subset <math>\{4\}</math> or <math>\{5\}</math>] = 8 - 8 = <math>0</math>
+If <math>n=2</math>, <math>\binom{8}{2}</math> - (<math>\binom{5}{2}</math> + <math>\binom{5}{2}</math> - 1) [subtract 1 to eliminate the overcounting of the subset <math>\{4,5\}</math>] = 28 - 19 = <math>9</math>
+If <math>n=3</math>, <math>\binom{8}{3}</math> - (<math>\binom{5}{3}</math> + <math>\binom{5}{3}</math>) = 56 - 20 = <math>36</math>
+If <math>n=4</math>, <math>\binom{8}{4}</math> - (<math>\binom{5}{4}</math> + <math>\binom{5}{4}</math>) = 70 - 10 = <math>60</math>
+If <math>n=5</math>, <math>\binom{8}{5}</math> - (<math>\binom{5}{5}</math> + <math>\binom{5}{5}</math>) = 56 - 2 = <math>54</math>
+If <math>n>5</math>, then the set <math>\{1,2,3,4,5,6,7,8\}</math> is never in violation of the two subsets <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math>. Thus,
+If <math>n=6</math>, <math>\binom{8}{6}</math> = <math>28</math>
+If <math>n=7</math>, <math>\binom{8}{7}</math> = <math>8</math>
+If <math>n=8</math>, <math>\binom{8}{8}</math> = <math>1</math>
+Adding these together, our solution becomes <math>0</math> + <math>9</math> + <math>36</math> + <math>60</math> + <math>54</math> + <math>28</math> + <math>8</math> + <math>1</math> <math>=\boxed{196}</math>
+Solution by IronicNinja~
 =See Also=
 {{AIME box|year=2017|n=II|before=First Problem|num-a=2}}
 {{MAA Notice}}

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