How many 3 digit numbers are there, such that the hundreds digit is strictly smaller than the tens digit, which is strictly smaller than the units digit?
Can you solve this without listing?
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Great! Thanks for highlighting this approach. In the previous question, most people did it the tedious way of listing out every case.
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My pleasure. :) If the order were strictly descending then we would have to bring $0$ back into play. Both problems would then become quite a bit harder if the order were no longer "strict".
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I believe that it's still just as "easy". You just have to find the correct bijection. Here is the question .
In the $10 \leq 1$ case, the answer is $45 = {1 0 \choose 2 }$ .
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@Chung Kevin – You're right: for some reason I remembered it as being a bit harder. I guess I was thinking of counting the number of, say, $10$ digit numbers whose digits were in (not strictly) ascending order. The method is still the same as the one I posted in your next problem, but it does get quite tedious.
Edit: At the prompting of Calvin, I found that there is a "simple" solution for (not strictly ascending) $n$ digit numbers in general, which I've outlined in a comment under the solution I posted to your next question.
Excellent solution.
What is the 'bracket' means? How it comes to 84 easily?
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The number of combinations of $r$ items chosen from a set of $n$ distinct items is $\dbinom{n}{r} = \dfrac{n!}{r!(n  r)!}.$
In this case we have $n = 9, r = 3,$ with
$\dbinom{9}{3} = \dfrac{9!}{3!*6!} = \dfrac{9*8*7}{6} = 3*4*7 = 84.$
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Oh it's another way to write 9C3. Great, really conscientious. Thanks.
I feel like I'm misunderstanding the question. If I have a 9 in the unit position, then I can have 2 thru 8 in the tens position. If it is a 2 in the tens position, then I can only have a 1 in the hundreds position. If it is an 8 in the tens position, then I can have 1 thru 7 in the hundreds position. Or not?
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Yes, that's the idea. The question is then how does one most efficiently count all such 3digit numbers, which is why I took the approach that I did and not count them case by case.
1 2 3 4 5 6 7 8 9 10 11 12 

That works too!
Since the hundreds digit cannot be zero and must be the smallest digit, we know that the 3digit numbers cannot have any zeroes. The set of numbers from which we can make 3digit numbers is therefor {1, 2, 3, 4, 5, 6, 7, 8, 9}.
To fill the three places (without repetition), we can pick from 9 numbers, then from 8 numbers, and finally from 7 numbers. The total number of 3digit numbers therefor is 9 x 8 x 7 = 504. But these do not contain only numbers that are ascending.
Let's say we pick three random distinct numbers from our set. For example, 2, 1, and 6. How many 3digit numbers can we make with only those numbers? Well, 3 x 2 x 1 = 6:
162 
216 
261 
612 
126 
621 
Notice that only one of them is ascending, 126. We need to eliminate 6 answers we're not looking for and that leaves us with the correct answer, 1. In other words, divide 6 by 6 to get 1.
For every 3digit number in the 504 3digit numbers, there are 6 versions of it, but in a different order. We need to get rid of those. In other words, divide 504 by 6 to get 84.
Thus its 9C3
If with think of a threedigit number as $i j k$ , then we see $\sum_{i=1}^7 \sum_{j=i+1}^8 \sum_{k=j+1}^9 1= 84$
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Since the hundreds digit cannot be $0$ and must be the smallest digit, we know that the $3$ digit numbers cannot have any $0$ 's.
Now for any three distinct digit chosen from the set $\{1,2,3,4,5,6,7,8,9\}$ there is only one way to arrange them in strictly ascending order, and there is onetoone correspondence between digits chosen in this manner and the $3$ digit numbers we are wanting to count. Thus the desired solution is $\dbinom{9}{3} = \boxed{84}.$