Difference between revisions of "2006 AMC 12A Problems/Problem 25"
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<math> \mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405</math> | <math> \mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405</math> | ||
| − | == Solution == | + | == Solution 1== |
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: | 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 <math>k</math> elements from an ordered <math>n</math> element [[set]] without choosing two consecutive members? | How many ways are there to choose <math>k</math> elements from an ordered <math>n</math> element [[set]] without choosing two consecutive members? | ||
| − | + | You want to choose <math>k</math> numbers out of <math>n</math> with no consecutive numbers. For each configuration, we can subtract <math>i-1</math> from the <math>i</math>-th element in your subset. This converts your configuration into a configuration with <math>k</math> elements where the largest possible element is <math>n-k+1</math>, with no restriction on consectutive numbers. Since this process is easily reversible, we have a [[bijection]]. | |
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Without consideration of the second condition, we have: | Without consideration of the second condition, we have: | ||
<math>{15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}</math> | <math>{15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}</math> | ||
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Now we have, after we cancel all the terms <math>{n \choose k}</math> where <math>n < k</math>, | Now we have, after we cancel all the terms <math>{n \choose k}</math> where <math>n < k</math>, | ||
<math>{15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}= 15 + 78 + 165 + 126 + 21 = \boxed{405} \Longrightarrow \mathrm{(E)}</math> | <math>{15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}= 15 + 78 + 165 + 126 + 21 = \boxed{405} \Longrightarrow \mathrm{(E)}</math> | ||
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| + | ==Solution 2== | ||
| + | We have the same setup as in the previous solution. | ||
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| + | Note that if <math>n < 2k - 1</math>, the answer will be 0. Otherwise, the <math>k</math> elements we choose define <math>k + 1</math> boxes into which we can drop the <math>n - k</math> remaining elements, with the caveat that each of the middle <math>k - 1</math> boxes must have at least one element. This is equivalent to dropping <math>n - 2k + 1</math> elements into <math>k + 1</math> boxes, where each box is allowed to be empty. And this is equivalent to arranging <math>n - k + 1</math> objects, <math>k</math> of which are dividers, which we can do in <math>F(n, k) = {n - k + 1 \choose k}</math> ways. | ||
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| + | Now, looking at our original question, we see that the thing we want to calculate is just <math>F(15, 1) + F(14, 2) + F(13, 3) + F(12, 4) + F(11, 5) = {15\choose 1} + {13\choose2} + {11\choose 3} + {9\choose 4} + {7 \choose 5} = 15 + 78 + 165 + 126 + 21 = 405 \Longrightarrow \mathrm{(E)}</math> | ||
== See also == | == See also == | ||
Revision as of 20:10, 17 February 2012
Contents
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 1
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?
You want to choose numbers out of with no consecutive numbers. For each configuration, we can subtract 
from the 
-th element in your subset. This converts your configuration into a configuration with elements where the largest possible element is 
, with no restriction on consectutive numbers. Since this process is easily reversible, we have a bijection.
Without consideration of the second condition, we have:
Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than , which translates to subtracting from the "top" of each binomial coefficient.
Now we have, after we cancel all the terms 
where 
,
Solution 2
We have the same setup as in the previous solution.
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 | |
