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p(Ascending Number) closest to .01 (Posted on 2020-03-16) Difficulty: 2 of 5
An Ascending Number has been defined by the following 2 properties:

- The number has at least 2 digits.

- All digits of the number are in strictly ascending order, in base 10.

For completeness, let's also add: no leading zeros or duplicate digits.

For what value of N is the probability that a randomly chosen N-digit number is also an ascending number closest to .01, i.e. a 1% chance?

For this value of N, what is the exact probability the random number N digits long is ascending?

No Solution Yet Submitted by Larry    
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Solution the full set | Comment 2 of 3 |
The number of possible random n-digit natural numbers is

9*10^(n-1)

To be ascending with no leading zeros means there are no zeros at all. As a result n can be no higher than 9 as duplicates are not allowed.

The choice of what digits are included in the strictly ascending number determines what that number is, as they must be arranged in order. So the cardinality of that set of numbers is C(9,n).

The required probability is therefore

C(9,n) / (9*10^(n-1))

The probabilities are:

 n   probability
 2   0.4 
 3   0.0933333333333333333 
 4   0.014 
 5   0.0014 
 6   0.0000933333333333332 
 7   0.000004 
 8   0.0000001 
 9   0.0000000011111111111 
 
The closest probability to 0.01 is for n=4:

126 / 9000 = 7/500 = .014
 
The above assumes the random number has no leading zeros but is allowed to have duplicate digits, as that rule applies only to the ascending number, making it strictly ascending. 

    5   open "probasn.txt" for output as #2
   10   for N=2 to 9
   20   print N,combi(9,N)/(9*10^(N-1))
   30   print #2,N,combi(9,N)/(9*10^(N-1))
  100   next
  110   close #2


  Posted by Charlie on 2020-03-16 15:08:07
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