Scanning a number from left to right, if the next digit is greater in value than the preceeding digit, we say that the number is strictly increasing: 1589, 3679, and 1348 are examples of 4-digit numbers with this property.
Given a certain number is strictly increasing, what is the probability that it contains
a. exactly 6 digits?
b. 6 digits or more?
Let's assume that we are only talking about integers.
Line the digits 1 through 9 up and form an increasing number by striking out 0 to 9 digits.
There are 2^9 ways of doing this, but let's also rule out 0 and 1 digit numbers.
Then the total number of valid increasing numbers are 2^9 - 10 = 512 - 10 = 502
Part a) To form a 6 digit number, strike out 3 of the 9 digits. The number of ways of doing this is C(9,3) = 9*8*7/3*2*1 = 84.
so the desired probability = 84/502 = 42/251, which is approximately 1/6
Part b) (84+36+9+1)/502 = 130/502 = 65/152
If we allow non-integers, then the probability tilts in favor of larger numbers, because 12 can be 12 or 1.2 or .12 (three possibilities) whereas a 6 digit number has seven possibilities. Since this is D2, I guess that this is not what the author had in mind.
Counting, Counting (spoiler)
