AIME 2004 I · 第 6 题

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 former case, suppose we pick digits $x_1,x_2,x_3,x_4$ such that $x_1. There are five arrangements of these digits that satisfy the condition of being snakelike:$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 $0. There are three arrangements of these digits that satisfy the condition of being snakelike:$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