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

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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 a_1 < a_2 < a_3 < a_4 < a_5 14</math>. If no
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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>b4 = a4 - 3</math>, and <math>b_5 = a_5 - 4</math> are distinct numbers from the interval [1, 10].
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<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 a1 = b1, a2 = b2 + 1, a3 = b3 + 2, a4 = b4 + 3, and
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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
a5 = b5+4 are from the interval [1, 14], and no two of them are consecutive.
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<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}
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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 750.
+
the answer is <math>750</math>.
 +
 
 +
== Solution 3 ==
 +
We will approach this problem using complementary counting. First it is obvious, there are <math>\binom{14}{5}</math> total sets. To find the number of sets with NO consecutive elements, we do a little experimentation. Consider the following:
 +
 
 +
Start of with any of the <math>14</math> numbers, say WLOG, <math>1</math>. Then we cannot have <math>2</math>. So again WLOG, we may pick <math>3</math>. Then we cannot have <math>4</math>, so again WLOG, we may pick <math>5</math>. Then not <math>6</math>, but <math>7</math>, then not <math>8</math>, but <math>9</math>. Now we have the set <math>{1,3,5,7,9}</math>, and we had to remove <math>4</math> digits from the <math>14</math> total amount, so there wasn't any consecutive numbers. So we have that the number of non-consecutive cardinality <math>5</math> sets, is <math>\binom{14-4}{5}=\binom{10}{5}</math>. Now you already read the solutions above, and if you are here, you are either confused, or looking for more. So now the answer is simply <math>1750</math>, which is <math>\boxed{750}</math> (mod <math>1000</math>).
 +
 
 +
~th1nq3r
 +
 
 +
(To get a feel for my solution, draw the set {_, _, _, _, _}, list all the <math>14</math> numbers on the side, and fill up the set starting with <math>1</math> and doing the steps I used for my solution).
  
 
== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2009|n=II|num-b=5|num-a=7}}
 +
[[Category: Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:47, 14 July 2021

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 $750$.

Solution 3

We will approach this problem using complementary counting. First it is obvious, there are $\binom{14}{5}$ total sets. To find the number of sets with NO consecutive elements, we do a little experimentation. Consider the following:

Start of with any of the $14$ numbers, say WLOG, $1$. Then we cannot have $2$. So again WLOG, we may pick $3$. Then we cannot have $4$, so again WLOG, we may pick $5$. Then not $6$, but $7$, then not $8$, but $9$. Now we have the set ${1,3,5,7,9}$, and we had to remove $4$ digits from the $14$ total amount, so there wasn't any consecutive numbers. So we have that the number of non-consecutive cardinality $5$ sets, is $\binom{14-4}{5}=\binom{10}{5}$. Now you already read the solutions above, and if you are here, you are either confused, or looking for more. So now the answer is simply $1750$, which is $\boxed{750}$ (mod $1000$).

~th1nq3r

(To get a feel for my solution, draw the set {_, _, _, _, _}, list all the $14$ numbers on the side, and fill up the set starting with $1$ and doing the steps I used for my solution).

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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