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 Pandigital + (Posted on 2017-05-03)
Let p(n) be the probability that a random n-digit integer has all 10 digits occurring each at least once.
Let us allow leading zeroes; e.g. 0000018274 will be included in the set of 10-digit numbers).

So p(9) is 0 , p(10) = 10!/1010 and p(1000) would be very close to 1.

a. What is the smallest n for which p(n) > 1/2?
b. >3/4?
c. >.95?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Matrix solution | Comment 2 of 3 |
I am going to redefine P(n) as a vector representing the frequencies of the number of different digits.  The elements in order are the fraction of n digit numbers with 1,2,3,...,10 different digits.  For example P(3) = [0.01, 0.27, 0.72, 0, 0, 0, 0, 0, 0, 0].  The tenth (last) element of the vector is the value discussed in the problem.

Let matrix T be a transition matrix so that T*P(n-1) = P(n).  P is being treated as a column vector in the multiplication.  Then T is
`[0.1  0   0   0   0   0   0   0   0   0 ][0.9 0.2  0   0   0   0   0   0   0   0 ][ 0  0.8 0.3  0   0   0   0   0   0   0 ][ 0   0  0.7 0.4  0   0   0   0   0   0 ][ 0   0   0  0.6 0.5  0   0   0   0   0 ][ 0   0   0   0  0.5 0.6  0   0   0   0 ][ 0   0   0   0   0  0.4 0.7  0   0   0 ][ 0   0   0   0   0   0  0.3 0.8  0   0 ][ 0   0   0   0   0   0   0  0.2 0.9  0 ][ 0   0   0   0   0   0   0   0  0.1  1 ]`
Then T^(n-1) * P(1) = P(n).  P(1) = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0], which means that P(n) is simple the first column of T^(n-1), and the value sought in the problem is the last element in that column.  Then finding the desired values is as simple as raising T to various powers and inspecting element T(10,1).

Then I find n=26 is the smallest n with T(10,1)>0.5, n=34 is the smallest with T(10,1)>0.75, and n=50 is the smallest with T(10,1)>0.95.  This means the answers to the the question for part a is 27, part b is 35, and part c is 51

(The best part is that I don't have to go through a ten level deep series of inclusion/exclusion probabilities, just plug a matrix expression in a suitable program and see what the result is.)

 Posted by Brian Smith on 2017-05-04 09:57:29

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