Difference between revisions of "2006 AMC 12A Problems/Problem 25"
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<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> | ||
− | 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 <math>k</math>, which translates to subtracting <math>k</math> from the "top" of each [[binomial coefficient]]. | + | 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 <math>k</math>, which translates to subtracting <math>k - 1</math> from the "top" of each [[binomial coefficient]]. |
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> |
Revision as of 17:43, 24 December 2018
Contents
[hide]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 consecutive 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
Another way of visualizing the solution above would be to use 's and
's. Denote
as the numbers we have chosen and
as other numbers. Taking an example, assuming we are picking two numbers, we imagine the shape
. This notation forces a number between the two chosen numbers, which blocks the two numbers we picked from being consecutive. Now we consider the orientations with this shape. We have
remaining numbers.
We need to find the number of ways to place the remaining 's. We can find this by utilizing stars and bars, with the following marker being placed to represent groups: *| - *|*. Now, we have to place
numbers within
groups, which is
. The same concept can be used for the remaining numbers. The rest of the solution continues as above.
Solution by: Everyoneintexas
Solution 3
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 (which divide the nonconsecutive numbers) into which we can drop the
remaining elements, with the caveat that each of the middle
boxes must have at least one element (since the numbers are nonconsecutive). 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.