100 < 10 < 1
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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4 solutions
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 1 0 ≤ 1 case, the answer is 4 5 = ( 2 1 0 ) .
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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, 1 0 -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 ( r n ) = r ! ( n − r ) ! n ! .
In this case we have n = 9 , r = 3 , with
( 3 9 ) = 3 ! ∗ 6 ! 9 ! = 6 9 ∗ 8 ∗ 7 = 3 ∗ 4 ∗ 7 = 8 4 .
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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 3-digit 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 |
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That works too!
Since the hundreds digit cannot be zero and must be the smallest digit, we know that the 3-digit numbers cannot have any zeroes. The set of numbers from which we can make 3-digit 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 3-digit 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 3-digit 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 3-digit number in the 504 3-digit 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 three-digit number as i j k , then we see ∑ i = 1 7 ∑ j = i + 1 8 ∑ k = j + 1 9 1 = 8 4
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 one-to-one correspondence between digits chosen in this manner and the 3 -digit numbers we are wanting to count. Thus the desired solution is ( 3 9 ) = 8 4 .