Answer:
252
Explanation:
The question is fully equivalent to asking how many subsets of length 5 there are of 10 objects (digits 0–9). That number is 10C5, where nCk is the number of ways to choose k objects from a list of 10. The value of that is ...
nCk = n!/(k!(n-k)!)
There are 252 ways to choose 5 numbers from the digits 0-9:
10C5 = 10!/(5!(10-5)!) = 10·9·8·7·6/(5·4·3·2·1) = 9·4·7 = 252
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The order of the selection doesn't matter, because the selected digits are always arranged in decreasing order to form a decreasing number.