Difference between revisions of "2017 AIME II Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
− | Upon inspection, a viable set must contain at least one element from both of the sets <math>\{1, 2, 3, 4, 5\}</math> and <math>\{4, 5, 6, 7, 8\}</math>. 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 <math>2^3=8</math>, but we want to exclude the empty set, giving us 7 ways to choose from <math>\{1, 2, 3\}</math> or <math>\{ | + | Upon inspection, a viable set must contain at least one element from both of the sets <math>\{1, 2, 3, 4, 5\}</math> and <math>\{4, 5, 6, 7, 8\}</math>. 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 <math>2^3=8</math>, but we want to exclude the empty set, giving us 7 ways to choose from <math>\{1, 2, 3\}</math> or <math>\{6, 7, 8\}</math>. We can take each of these <math>7 \times 7=49</math> 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 <math>49 \times 4=\boxed{196}</math>. |
==Solution 3 == | ==Solution 3 == | ||
− | The set of all subsets of <math>\{1,2,3,4,5,6,7,8\}</math> that are disjoint with respect to <math>\{4,5\}</math> and are not disjoint with respect to the complements of sets (and therefore not a subset of) <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math> will be named <math>S</math>, which has <math>7\cdot7=49</math> members. The union of each member in <math>S</math> and the <math>2^2=4</math> subsets of <math>\{4,5\}</math> will be the members of set <math>Z</math>, which has <math>49\cdot4=\boxed{196}</math> members. <math>\blacksquare</math> | + | This solution is very similar to Solution <math>2</math>. The set of all subsets of <math>\{1,2,3,4,5,6,7,8\}</math> that are disjoint with respect to <math>\{4,5\}</math> and are not disjoint with respect to the complements of sets (and therefore not a subset of) <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math> will be named <math>S</math>, which has <math>7\cdot7=49</math> members. The union of each member in <math>S</math> and the <math>2^2=4</math> subsets of <math>\{4,5\}</math> will be the members of set <math>Z</math>, which has <math>49\cdot4=\boxed{196}</math> members. <math>\blacksquare</math> |
Solution by [[User:a1b2|a1b2]] | 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>-</math> (<math>\binom{5}{1}</math> + <math>\binom{5}{1}</math> <math>- 2</math>) [subtract <math>2</math> to eliminate the overcounting of the subset <math>\{4\}</math> or <math>\{5\}</math>] = <math>8 - 8</math> = <math>0</math> | ||
+ | |||
+ | If <math>n=2</math>, <math>\binom{8}{2}</math> <math>-</math> (<math>\binom{5}{2}</math> + <math>\binom{5}{2}</math> <math>- 1</math>) [subtract <math>1</math> to eliminate the overcounting of the subset <math>\{4,5\}</math>] = <math>28 - 19</math> = <math>9</math> | ||
+ | |||
+ | If <math>n=3</math>, <math>\binom{8}{3}</math> <math>-</math> (<math>\binom{5}{3}</math> + <math>\binom{5}{3}</math>) = <math>56 - 20</math> = <math>36</math> | ||
+ | |||
+ | If <math>n=4</math>, <math>\binom{8}{4}</math> <math>-</math> (<math>\binom{5}{4}</math> + <math>\binom{5}{4}</math>) = <math>70 - 10</math> = <math>60</math> | ||
+ | |||
+ | If <math>n=5</math>, <math>\binom{8}{5}</math> <math>-</math> (<math>\binom{5}{5}</math> + <math>\binom{5}{5}</math>) = <math>56 - 2</math> = <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= | =See Also= | ||
{{AIME box|year=2017|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2017|n=II|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:02, 29 January 2024
Problem
Find the number of subsets of that are subsets of neither
nor
.
Solution 1
The number of subsets of a set with elements is
. The total number of subsets of
is equal to
. The number of sets that are subsets of at least one of
or
can be found using complementary counting. There are
subsets of
and
subsets of
. It is easy to make the mistake of assuming there are
sets that are subsets of at least one of
or
, but the
subsets of
are overcounted. There are
sets that are subsets of at least one of
or
, so there are
subsets of
that are subsets of neither
nor
.
.
Solution 2
Upon inspection, a viable set must contain at least one element from both of the sets and
. 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
, but we want to exclude the empty set, giving us 7 ways to choose from
or
. We can take each of these
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
.
Solution 3
This solution is very similar to Solution . The set of all subsets of
that are disjoint with respect to
and are not disjoint with respect to the complements of sets (and therefore not a subset of)
and
will be named
, which has
members. The union of each member in
and the
subsets of
will be the members of set
, which has
members.
Solution by a1b2
Solution 4
Consider that we are trying to figure out how many subsets are possible of that are not in violation of the two subsets
and
. Assume that the number of numbers we pick from the subset
is
. Thus, we can compute this problem with simple combinatorics:
If ,
(
+
) [subtract
to eliminate the overcounting of the subset
or
] =
=
If ,
(
+
) [subtract
to eliminate the overcounting of the subset
] =
=
If ,
(
+
) =
=
If ,
(
+
) =
=
If ,
(
+
) =
=
If , then the set
is never in violation of the two subsets
and
. Thus,
If ,
=
If ,
=
If ,
=
Adding these together, our solution becomes +
+
+
+
+
+
+
Solution by IronicNinja~
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.