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Difference between revisions of "2004 AIME I Problems/Problem 6"

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In the second case we choose zero and three other digits such that <math>0<x_2<x_3<x_4</math>. There are three arrangements of these digits that satisfy the condition of being snakelike: <math>x_2x_30x_4</math>, <math>x_2x_40x_3</math>, <math>x_3x_40x_2</math>. Because we know that zero is a digit, there are <math>3\cdot{9\choose 3}=252</math> snakelike numbers which contain the digit zero. Thus there are <math>630+252=\boxed{882}</math> snakelike numbers.
 
In the second case we choose zero and three other digits such that <math>0<x_2<x_3<x_4</math>. There are three arrangements of these digits that satisfy the condition of being snakelike: <math>x_2x_30x_4</math>, <math>x_2x_40x_3</math>, <math>x_3x_40x_2</math>. Because we know that zero is a digit, there are <math>3\cdot{9\choose 3}=252</math> snakelike numbers which contain the digit zero. Thus there are <math>630+252=\boxed{882}</math> snakelike numbers.
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+
=
 
== Solution 2 ==
 
== Solution 2 ==
 
Let's create the snakelike number from digits <math>a < b < c < d</math>, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of <math>5\cdot{10 \choose 4}</math> But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits <math>a < b < c</math>. There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is <math>2\cdot{9 \choose 3}</math>. Thus our answer is <math>5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}</math>
 
Let's create the snakelike number from digits <math>a < b < c < d</math>, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of <math>5\cdot{10 \choose 4}</math> But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits <math>a < b < c</math>. There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is <math>2\cdot{9 \choose 3}</math>. Thus our answer is <math>5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}</math>
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 +
== Solution 3 ==
 +
We will first decide the order of the 4 digits, greatest to least. To do this, we will pretend that we have selected the digits 1,2,3,4, and we need to arrange them to create a snakelike number. By testing all permutations, there are only 5 ways to make a snakelike number: (1,3,2,4),(1,4,2,3),(2,3,1,4),(2,4,1,3),(3,4,1,2).
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 +
Now we select 4 digits to replace the 1,2,3,4.
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 +
In first 2 of cases: (1,3,2,4),(1,4,2,3), the leading digit is a 1, which means it is the smallest of our 4 digits. If we select a 0, the leading digit will be a zero, which is bad because all numbers between 1000 and 9999 have nonzero leading digits. So, we need to select our 4 digits only from the pool of 1-9. There are 9 choose 4 ways and there are 2 cases:
 +
 +
<math>2 \cdot \dbinom{9}{4} = 252</math>
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 +
Thus, there are 252 ways for those 2 cases.
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For the next 3 cases, selecting a 0 is okay, so we can select from the pool of 0-9. There are 10 choose 4 ways to select our 4 digits and there are 3 cases:
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<math>3 \cdot \dbinom{10}{4} = 630</math>
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For those 3 cases there are 630 ways.
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Thus, our answer is 630+252 = <math>\boxed{882}</math>.
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-Alexlikemath
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2004|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2004|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:59, 14 March 2020

Problem

An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ satisfies $a_i<a_{i+1}$ if $i$ is odd and $a_i>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits?

Solution 1

We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$. There are five arrangements of these digits that satisfy the condition of being snakelike: $x_1x_3x_2x_4$, $x_1x_4x_2x_3$, $x_2x_3x_1x_4$, $x_2x_4x_1x_3$, $x_3x_4x_1x_2$. Thus there are $5\cdot {9\choose 4}=630$ snakelike numbers which do not contain the digit zero.

In the second case we choose zero and three other digits such that $0<x_2<x_3<x_4$. There are three arrangements of these digits that satisfy the condition of being snakelike: $x_2x_30x_4$, $x_2x_40x_3$, $x_3x_40x_2$. Because we know that zero is a digit, there are $3\cdot{9\choose 3}=252$ snakelike numbers which contain the digit zero. Thus there are $630+252=\boxed{882}$ snakelike numbers. =

Solution 2

Let's create the snakelike number from digits $a < b < c < d$, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of $5\cdot{10 \choose 4}$ But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits $a < b < c$. There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is $2\cdot{9 \choose 3}$. Thus our answer is $5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}$

Solution 3

We will first decide the order of the 4 digits, greatest to least. To do this, we will pretend that we have selected the digits 1,2,3,4, and we need to arrange them to create a snakelike number. By testing all permutations, there are only 5 ways to make a snakelike number: (1,3,2,4),(1,4,2,3),(2,3,1,4),(2,4,1,3),(3,4,1,2).

Now we select 4 digits to replace the 1,2,3,4.

In first 2 of cases: (1,3,2,4),(1,4,2,3), the leading digit is a 1, which means it is the smallest of our 4 digits. If we select a 0, the leading digit will be a zero, which is bad because all numbers between 1000 and 9999 have nonzero leading digits. So, we need to select our 4 digits only from the pool of 1-9. There are 9 choose 4 ways and there are 2 cases:

$2 \cdot \dbinom{9}{4} = 252$

Thus, there are 252 ways for those 2 cases.

For the next 3 cases, selecting a 0 is okay, so we can select from the pool of 0-9. There are 10 choose 4 ways to select our 4 digits and there are 3 cases:

$3 \cdot \dbinom{10}{4} = 630$

For those 3 cases there are 630 ways.

Thus, our answer is 630+252 = $\boxed{882}$.

-Alexlikemath

See also

2004 AIME I (Problems • Answer Key • Resources)
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