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seen Jun 11 '13 at 7:57

Jan
24
awarded  Editor
Jan
24
comment How many ways can 1's and 2's be added to equal 17 if order matters?
Okay, I've done that and figured out with Google that the pattern you refer to is the Fibonacci pattern. Unfortunately we haven't learned about that in our coursework, so I'm not comfortable basing my answer on it. I'm still working on the proof aspect though, since that is something I could use.
Jan
24
revised How many ways can 1's and 2's be added to equal 17 if order matters?
edited title
Jan
24
asked How many ways can 1's and 2's be added to equal 17 if order matters?
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24
awarded  Student
Jan
24
awarded  Scholar
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
Got the answer of 336 and I'm pretty sure I understand why. My brain is resisting it for some reason, but I know it's right and I could do it again, so you get credit for the answer. I would vote it up if I had the rep for it, so instead accept my thanks for being patient with me through that.
Jan
24
accepted How many ascending and descending numbers are between 1000 and 9999?
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
Okay, something clicked that time. So, I'm going to take 4 numbers out of the nine, but I'm going to do this twice. Then, I have to account for the possible zeroes when descending, so I'm going to assume a zero at the end, which means I'll be taking 3 of the 9 numbers, which will be added to the other two where I took 4. And here's my last shot at notation. =) EDIT: Notation didn't work out, it was formatted away. Imagine the numbers in parenthesis vertical: 2(9 4)+(9 3)
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
Better notation to use there might be P(9,4) for ascending, as it's how I usually see nPr written when formula editors are not available, but it could also be written as 9P4 I suppose. So lets say I choose 1234. They are already arranged as ascending. So I use P(9,4). It seems to me that this wouldn't include combinations such as, say, 5678. As I said in reply to one of the comments above, there is something very basic that my mind hasn't accepted yet.
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
@JulianKuelshammer - I'll give that a try. That has worked for me on some of the other permutation puzzles we've been doing, but I didn't think it would offer any insight on this one. This is a much more difficult problem than the others due to some base concept I seem to be missing.
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
@DougSpoonwood - Okay, I think I see what you mean, I would have 10 choices there, unless I wasn't counting the zeroes for the ascending group. I don't quite understand how x, y, and z are known though.
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
Okay, 4 digits from 10. Well, ascending it would be four digits from 9 since no zeroes work. Descending will work with zeroes, though, so that could be 4 digits from 10. So it seems maybe I'm looking at (9 4) + (10 4)? Also, I really don't understand how this could work this way. How does solving for choosing four digits get me the number of compositions that are only ascending?
Jan
24
comment How many ascending and descending numbers are between 1000 and 9999?
Hmm, that's a good nudge for me, and helps with part of what I was confused about, but without considering the 0 for increasing permutations I see it as P(9000, 4), which I KNOW isn't right. 9000! would be an absolutely ridiculous number such that dividing it by 4! wouldn't bring it anywhere near the range I'm looking for.
Jan
24
asked How many ascending and descending numbers are between 1000 and 9999?