Difference between revisions of "2006 AMC 12A Problems/Problem 25"
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<math>(2)</math> If <math>S</math> contains <math>k</math> [[element]]s, then <math>S</math> contains no number less than <math>k</math>. | <math>(2)</math> If <math>S</math> contains <math>k</math> [[element]]s, then <math>S</math> contains no number less than <math>k</math>. | ||
| − | <math> \mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377 | + | <math> \mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405</math> |
== Solution == | == Solution == | ||
Revision as of 12:45, 13 December 2007
Problem
How many non- empty subsets 
of 
have the following two properties?
No two consecutive integers belong to .
If contains 
elements, then contains no number less than .
Solution
This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:
How many ways are there to choose elements from an ordered 
element set without choosing two consecutive members?
Note that if 
, the answer will be 0. Otherwise, the elements we choose define 
boxes into which we can drop the 
remaining elements, with the caveat that each of the middle 
boxes must have at least one element. This is equivalent to dropping 
elements into boxes, where each box is allowed to be empty. And this is equivalent to arranging 
objects, of which are dividers, which we can do in 
ways.
Now, looking at our original question, we see that the thing we want to calculate is just 
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Question |
| 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 | |
