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How many 3-digit numbers between 100 and 999 are there with the property that each digit, from the left, beginning with the 2nd is > than the digit before it, and the 3rd is > than the 2nd, such as : 123, 234, 456....etc? And how many more 6-digit such numbers are there between 100,000 and 999,999? Any help would be great. Thank you. 

 May 15, 2019
 #1
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How many 3-digit numbers between 100 and 999 are there with the property that each digit, from the left, beginning with the 2nd is > than the digit before it, and the 3rd is > than the 2nd, such as : 123, 234, 456....etc? And how many more 6-digit such numbers are there between 100,000 and 999,999? Any help would be great. Thank you. 

 

It stands to reason that 0 cannot be one of the digits and all the digits must be different. 

How many ways can 3 digits be chosen from 9, that would be 9C3=84

For each of those triples there is only one way that they can be placed in ascending order

So the answer is 84

 

You can now work out the number of 6 digit combinations for yourself.  It will be done in the same way.

 May 15, 2019
 #2
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Thank you very much.

 May 15, 2019

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