Difference between revisions of "2023 AMC 12B Problems/Problem 8"
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==Problem== | ==Problem== | ||
| − | 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. | + | 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> | ||
Latest revision as of 19:51, 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. 
