Difference between revisions of "2009 AIME II Problems/Problem 6"
| Line 9: | Line 9: | ||
Hence we have <math>m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750</math>, and the answer is <math>1750 \bmod 1000 = \boxed{750}</math>. | Hence we have <math>m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750</math>, and the answer is <math>1750 \bmod 1000 = \boxed{750}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | Let A be the number of ways in which 5 distinct numbers can be selected | ||
| + | from the set of the first 14 natural numbers, and let B be the number of | ||
| + | ways in which 5 distinct numbers, no two of which are consecutive, can | ||
| + | be selected from the same set. Then m = A − B. Because A = <math>\dbinom{14}{5}</math>, the | ||
| + | problem is reduced to finding B. | ||
| + | Consider the natural numbers <math>1 ≤ a_1 < a_2 < a_3 < a_4 < a_5 ≤ 14</math>. If no | ||
| + | two of them are consecutive, the numbers <math>b_1 = a_1, b_2 = a_2 - 1</math>, <math>b_3 = a_3 - 2</math>, | ||
| + | <math>b4 = a4 - 3</math>, and <math>b_5 = a_5 - 4</math> are distinct numbers from the interval [1, 10]. | ||
| + | Conversely, if <math>b_1 < b_2 < b_3 < b_4 < b_5</math> are distinct natural numbers from | ||
| + | the interval [1, 10], then a1 = b1, a2 = b2 + 1, a3 = b3 + 2, a4 = b4 + 3, and | ||
| + | a5 = b5+4 are from the interval [1, 14], and no two of them are consecutive. | ||
| + | Therefore counting B is the same as counting the number of ways of | ||
| + | choosing 5 distinct numbers from the set of the first 10 natural numbers. | ||
| + | Thus B = <math>\dbinom{10}{5}</math>. Hence m = A − B = <math>\dbinom{14}{5} - \dbinom{10}{5} | ||
| + | = 2002 - 252 = 1750</math> and | ||
| + | the answer is 750. | ||
== See Also == | == See Also == | ||
Revision as of 16:10, 26 October 2014
Contents
Problem
Let 
be the number of five-element subsets that can be chosen from the set of the first 
natural numbers so that at least two of the five numbers are consecutive. Find the remainder when is divided by 
.
Solution
We can use complementary counting. We can choose a five-element subset in 
ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 
s and 5 
s, thereby showing that there are 
such sets.
Given a five-element subset 
of 
in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the 
-th character is if 
and otherwise. Now we got a string with 5 s and 9 s. As no two numbers were consecutive, we know that in our string no two s are consecutive. We can now remove exactly one from between each pair of s to get a string with 5 s and 5 s. And clearly this is a bijection, as from each string with 5 s and 5 s we can reconstruct one original set by reversing the construction.
Hence we have 
, and the answer is 
.
Solution 2
Let A be the number of ways in which 5 distinct numbers can be selected
from the set of the first 14 natural numbers, and let B be the number of
ways in which 5 distinct numbers, no two of which are consecutive, can
be selected from the same set. Then m = A − B. Because A = 
, the
problem is reduced to finding B.
Consider the natural numbers $1 ≤ a_1 < a_2 < a_3 < a_4 < a_5 ≤ 14$ (Error compiling LaTeX. Unknown error_msg). If no
two of them are consecutive, the numbers 
, 
,
, and 
are distinct numbers from the interval [1, 10].
Conversely, if 
are distinct natural numbers from
the interval [1, 10], then a1 = b1, a2 = b2 + 1, a3 = b3 + 2, a4 = b4 + 3, and
a5 = b5+4 are from the interval [1, 14], and no two of them are consecutive.
Therefore counting B is the same as counting the number of ways of
choosing 5 distinct numbers from the set of the first 10 natural numbers.
Thus B = 
. Hence m = A − B = 
and
the answer is 750.
See Also
| 2009 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
