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I got the answer to the question: The number of possible $4$ digit numbers with ordered digits

But I wasn't sure if I could use the same principle outlined there to solve this:

How many possible 4 digit numbers are there where each digit is lower than or equal to the one before it?

This question would allow numbers such as 5332 which the previous question wouldn't.

How do I account for the extra solutions, past 210?

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Use stars-and-bars and relate your original problem to the problem of having ten labeled buckets (labeled 0,1,2,...9) and four identical balls.

For example, an outcome of two balls in the bucket labeled 5, one ball in the bucket labeled 7 and one ball in the bucket labeled 2 would correspond to the four digit number $7552$.

Recognize now that all of the balls-and-buckets outcomes make sense as your four digit numbers in decreasing order with a small exception.

The outcome of all balls in the bucket labeled 0 would correspond to the number "$0000$" which is not traditionally considered a four-digit number.

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  • $\begingroup$ Ok, using your method it seems to give me the right answer... Thank You $\endgroup$ – The Math Guy Nov 29 '17 at 3:19

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