Difference between revisions of "2023 AMC 12B Problems/Problem 8"
Scrabbler94 (talk | contribs) m (→Problem) |
|||
| (5 intermediate revisions by 5 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | How many nonempty subsets B of <math>{0, 1, 2, 3, \cdots, 12}</math> have the property that the number of elements in B is equal to the least element of B? For example, | + | How many nonempty subsets <math>B</math> of <math>\{0, 1, 2, 3, \cdots, 12\}</math> have the property that the number of elements in <math>B</math> is equal to the least element of <math>B</math>? For example, <math>B = \{4, 6, 8, 11\}</math> satisfies the condition. |
<math>\textbf{(A) } 256 \qquad\textbf{(B) } 136 \qquad\textbf{(C) } 108 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156</math> | <math>\textbf{(A) } 256 \qquad\textbf{(B) } 136 \qquad\textbf{(C) } 108 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156</math> | ||
==Solution 1== | ==Solution 1== | ||
| − | There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do <math>\binom{10}{1}</math>. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do <math>\binom{9}{2}</math>. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add <math>1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}</math>. This is <math>1+10+36+56+35+6 = \boxed{144}</math>. | + | There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do <math>\binom{10}{1}</math>. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do <math>\binom{9}{2}</math>. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add <math>1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}</math>. This is <math>1+10+36+56+35+6 = \boxed{\textbf{(D) 144}}</math>. |
~lprado | ~lprado | ||
| + | ==Note:== | ||
| + | In general, i.e. when the number is not <math>13</math>, the answer is <math>F_{n-1}</math>. | ||
| + | -Mr Sharkman | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/r9HCzaxLNlc | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2023|ab=B|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:51, 13 April 2024
Problem
How many nonempty subsets 
of 
have the property that the number of elements in is equal to the least element of ? For example, 
satisfies the condition.
Solution 1
There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do 
. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do 
. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add 
. This is 
.
~lprado
Note:
In general, i.e. when the number is not 
, the answer is 
.
-Mr Sharkman
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2023 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
