How many ascending and descending numbers are between 1000 and 9999? I'm working on some homework right now, and I've gotten stumped. Here's the question I'm on:
How many of the 9000 four-digit integers 1000, 1001, 1002, . . . , 9998, 9999 have four distinct digits that are either increasing (as in 1347 and 6789) or decreasing (as in 6421 and 8653)?
Through some Googling I've already found what they answer may be, but I have no idea how to arrive at the answer,  so I wouldn't be learning anything from this hurdle like I'm supposed to. I can use all the help I can get on this one since I don't even know where to start. I think I have to apply the formula for permutations to this somehow, but I don't know what adjustments I have to make for it to work right. 
 A: Once you've chosen the four digits, there's only one way to arrange them increasing, and one way decreasing, so it comes down to, how many ways are there to choose four digits? Oh, there's one little tricky bit: if you choose zero, you can't do increasing. 
A: If you're looking for how many 4 digit numbers are increasing or decreasing between 1000 and 9999, the answer has been provided here: How many of the 9000 four digit integers have four digits that are increasing?
If you (or since this was posted 3 years ago, if a discrete math student) are looking for how many non-decreasing or non-increasing 4-digit numbers there are between 1000 and 9999, I believe I have the answer.
*the difference is that non-decreasing can have 1226 or 7888 while increasing cannot have repetition. 
$\binom{10}{4}$ is the correct answer for decreasing combinations, while $\binom{10}{4} - {9 \choose 3}$ is the answer for increasing combinations because you can't begin the 4 digit segment with (or include) a zero.
The same idea applies to non-decreasing / non-increasing values, this time we allow for repetition, so for every integer after the 1st integer, we have +1 choices in digits. 
For decreasing numbers, 3210 was the only option where the 1st digit was 3. This is reflected by $\binom{3}{3}$. Now, since we can have 4-digit numbers like 3332, 3200, and 3111, we have $\binom{3+3}{3}$ combinations. We then have to subtract 1 because 3333 is not a non-increasing number. 
So, this same principle applies. We get $\binom{10+3}{4}$ + ($\binom{10+3} {4} - \binom{9+3}{3}$) - 10 = 1200. We subtract ten because 1111,2222,...9999 are not valid options for either non-increasing nor non-decreasing.
Final answer = 1200.
