Difference between revisions of "2009 AIME II Problems/Problem 6"

(Solution 2)
m (Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
  
Let A be the number of ways in which 5 distinct numbers can be selected
+
Let <math>A</math> be the number of ways in which <math>5</math> distinct numbers can be selected
from the set of the first 14 natural numbers, and let B be the number of
+
from the set of the first <math>14</math> natural numbers, and let <math>B</math> be the number of
ways in which 5 distinct numbers, no two of which are consecutive, can
+
ways in which <math>5</math> 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
+
be selected from the same set. Then <math>m = A -B</math>. Because <math>A = \dbinom{14}{5}</math>, the
problem is reduced to finding B.
+
problem is reduced to finding <math>B</math>.
 
Consider the natural numbers <math>1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 14</math>. If no
 
Consider the natural numbers <math>1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 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>,
 
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>b_4 = a_4 - 3</math>, and <math>b_5 = a_5 - 4</math> are distinct numbers from the interval [1, 10].
+
<math>b_4 = a_4 - 3</math>, and <math>b_5 = a_5 - 4</math> are distinct numbers from the interval <math>[1, 10]</math>.
 
Conversely, if <math>b_1 < b_2 < b_3 < b_4 < b_5</math> are distinct natural numbers from
 
Conversely, if <math>b_1 < b_2 < b_3 < b_4 < b_5</math> are distinct natural numbers from
the interval [1, 10], then <math>a_1 = b_1</math>, <math>a_2 = b_2 + 1</math>, <math>a_3 = b_3 + 2</math>, <math>a_4 = b_4 + 3</math>, and
+
the interval <math>[1, 10]</math>, then <math>a_1 = b_1</math>, <math>a_2 = b_2 + 1</math>, <math>a_3 = b_3 + 2</math>, <math>a_4 = b_4 + 3</math>, and
<math>a_5 = b_5 + 4</math> are from the interval [1, 14], and no two of them are consecutive.
+
<math>a_5 = b_5 + 4</math> are from the interval <math>[1, 14]</math>, and no two of them are consecutive.
Therefore counting B is the same as counting the number of ways of
+
Therefore counting <math>B</math> is the same as counting the number of ways of
choosing 5 distinct numbers from the set of the first 10 natural numbers.
+
choosing <math>5</math> distinct numbers from the set of the first <math>10</math> natural numbers.
Thus B = <math>\dbinom{10}{5}</math>. Hence m = A B = <math>\dbinom{14}{5} - \dbinom{10}{5}
+
Thus <math>B = \dbinom{10}{5}</math>. Hence <math>m = A - B = \dbinom{14}{5} - \dbinom{10}{5}
 
= 2002 - 252 = 1750</math> and
 
= 2002 - 252 = 1750</math> and
 
the answer is <math>750</math>.
 
the answer is <math>750</math>.

Revision as of 20:16, 13 March 2015

Problem

Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$.

Solution

We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ 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 $A$s and 5 $B$s, thereby showing that there are ${10\choose 5}$ such sets.

Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$-th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$s and 9 $B$s. As no two numbers were consecutive, we know that in our string no two $A$s are consecutive. We can now remove exactly one $B$ from between each pair of $A$s to get a string with 5 $A$s and 5 $B$s. And clearly this is a bijection, as from each string with 5 $A$s and 5 $B$s we can reconstruct one original set by reversing the construction.

Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$, and the answer is $1750 \bmod 1000 = \boxed{750}$.

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 $A = \dbinom{14}{5}$, the problem is reduced to finding $B$. Consider the natural numbers $1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 14$. If no two of them are consecutive, the numbers $b_1 = a_1, b_2 = a_2 - 1$, $b_3 = a_3 - 2$, $b_4 = a_4 - 3$, and $b_5 = a_5 - 4$ are distinct numbers from the interval $[1, 10]$. Conversely, if $b_1 < b_2 < b_3 < b_4 < b_5$ are distinct natural numbers from the interval $[1, 10]$, then $a_1 = b_1$, $a_2 = b_2 + 1$, $a_3 = b_3 + 2$, $a_4 = b_4 + 3$, and $a_5 = b_5 + 4$ 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 $B = \dbinom{10}{5}$. Hence $m = A - B = \dbinom{14}{5} - \dbinom{10}{5} = 2002 - 252 = 1750$ and the answer is $750$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
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

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