
#1
Nov209, 02:42 PM

P: 116

1. The problem statement, all variables and given/known data
How many four digit numbers exist where the a) the digits are strictily increasing (i.e. 1234 and 1579 but not 1338 or 1748) in size, b) decreasing in size? 2. Relevant equations 3. The attempt at a solution My problem is that I don't really have a good motivation for all this. The reasoning is something like "there are four places and nine numbers, which makes it the binominal coefficient nine over four", but it's a bit.. holey. And the guy that checks this is really strict. Anyway, that's the problem for the other one as well. I know it's the binomial coefficient 10 over 4, but how to motivate it? Tips, anyone? 



#2
Nov209, 03:29 PM

P: 740

Since you seem to have figured out the correct answer, you must have some sort of reasoning, even if it isn't clear to you right now. Here are a few questions you can answer to try to motivate your solution:
1. Explain why it is 9C4 for (a) but 10C4 for (b). 2. Explain why you used 9C4 and not, say, 9*10^{3}, which is how many 4 digit numbers there are in total. 3. Explain why you have used combinations and not permutations. 



#3
Nov309, 12:11 PM

P: 116

Ah, I figured it out :)
I choose four numbers, but there are 24 ways to combine those numbers and only one combination is possible, and therefore I've got 9*8*7*6/24 and since 4!=24, I can "add" 5! to both the denominator and nominator, which makes it 9C4. And it's 10 in b) because I can use 0 as well, since it can be in the last position and in a) I can't have it at all, since it'd have to be in the beginning and a natural number can't start with a 0. Anyway, thanks for the pointers, it made me think in a bit of a different way :) 


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