2004 AIME I Problems/Problem 6
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 such that . There are five arrangements of these digits that satisfy the condition of being snakelike: , , , , . Thus there are snakelike numbers which do not contain the digit zero.
In the second case we choose zero and three other digits such that . There are three arrangements of these digits that satisfy the condition of being snakelike: , , . Because we know that zero is a digit, there are snakelike numbers which contain the digit zero. Thus there are snakelike numbers. =
Let's create the snakelike number from digits , 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 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 . 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 . Thus our answer is
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:
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:
For those 3 cases there are 630 ways.
Thus, our answer is 630+252 = .
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