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

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== Solution 3 ==
 
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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:
 
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:
  

Revision as of 11:47, 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 (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

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