Ten Different Digits

What is the probability that a number chosen at random from the range 1,\!000,\!000,\!000 to 9,\!999,\!999,\!999 inclusive will contain ten different digits? Round your answer to the nearest millionth.
Source: NCTM Mathematics Teacher, January 2006

Solution
We divide the large range into nine smaller ranges:
1,\!000,\!000,\!000 to 1,\!999,\!999,\!999
2,\!000,\!000,\!000 to 2,\!999,\!999,\!999
3,\!000,\!000,\!000 to 3,\!999,\!999,\!999
\cdots
8,\!000,\!000,\!000 to 8,\!999,\!999,\!999
9,\!000,\!000,\!000 to 9,\!999,\!999,\!999

We examine the first range in details.
The first digit after  1 has nine choices \left \{0,2,3,4,5,6,7,8,9\right \}
The second digit after 1 has eight choices
The third digit after 1 has seven choices, etc.
Number of desirable outcomes
1\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 = 9!
Similarly, each of the other ranges has 9! desirable outcomes.
Total number of desirable outcomes = 9\times 9!
Total number of possible outcomes
9,\!999,\!999,\!999-1,\!000,\!000,\!000+1=9\times 10^9

Probability that a number chosen at random will contain ten different digits
\dfrac{9\times 9!}{9\times 10^9}=\dfrac{9!}{10^9}
=0.000363

Answer: 0.000363

 

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About mvtrinh

Retired high school math teacher.
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