What is Proper?

Since we are looking for $6$ digit numbers monotonically increasing, once we choose 6 digits out of $9$ $(1$ to $9)$ , there's just one way of arranging them. Let's say we choose the digits: ${7,1,1,3,4,4}$ . Once we order them, there's just ONE possible increasing order: $113447$ So our problem is finding the number of combinations we can make by choosing $6$ digits (with repetition) out of $9$ . The best way of doing this is imagining the amount of digits $(6)$ as dots and separate them into $9$ little homes (which correspond to the digits we are choosing out of $9$ $(1$ to $9)$ ). For example, the combination ${1,3,3,5,7,9}$ corresponds to: $\underbrace{*}_{1}\underbrace{||}_{2}\underbrace{**}_{3}\underbrace{||}_{4}\underbrace{*}_{5}\underbrace{||}_{6}\underbrace{*}_{7}\underbrace{||}_{8}\underbrace{*}_{9}$ It means we got: $1$ digit $1$ ; $0$ digit $2$ ; $2$ digits $3$ ; $0$ digit $4$ ; $1$ digit $5$ ; $0$ digit $6$ ; $1$ digit $7$ ; $0$ digit $8$ and $1$ digit $9$ . So in order to we calculate the total of proper numbers, we have to calculate the amount of permutations of: $******||||||||$ which is: ${{14}\choose{6}} = \boxed{3003}.$

(from the 14 places available, we choose 6 to put the dots $*$ )

Define $A_{i}=a_{i}+i-1$ where $a_{i}$ is the ith digit of the six-digit number, i=1, 2, 3, 4, 5, 6. Note that $A_{i}$ is strictly increasing when $i$ increases if and only if the six digit number satisfies the conditions in the question. Noting that $A_{i}$ can take any value from 1 to 9+5=14, the answer is just $^{14} C_{6}=3003$ .