Repeating digits

Let's see how many positive numbers from 1 up to and including 2016 have no repeating digits.

Let $d_n$ be the $n^{th}$ digit of the number.

1 digit numbers:

All 9 single digits numbers have no repeating digits.

2 digit numbers:

$d_1 \in \{1, 2, ..., 9\}$ and $d_2 \in \{0, 1, ..., 9\}$ but different from the first.

So a total of $9 \times 9 = 81$ two-digit numbers have no repeating digits.

3 digit numbers:

$d_1 \in \{1, 2, ..., 9\}$ and $d_2 \in \{0, 1, ..., 9\}$ but different from the first and $d_3 \in \{0, 1, ..., 9\}$ but different from the first and second.

So a total of $9 \times 9 \times 8 = 648$ three-digit numbers have no repeating digits.

4 digit numbers starting with 1:

$d_1 = 1$ and $d_2 \in \{0\} \cup \{2, ..., 9\}$ and $d_3 \in \{0\} \cup \{2, ..., 9\}$ but different from the second and $d_4 \in \{0\} \cup \{2, ..., 9\}$ but different from the second and the third.

So a total of $1 \times 9 \times 8 \times 7 = 504$ four-digit numbers have no repeating digits.

4 digit numbers starting with 2 (up to 2016):

Only 2013, 2014, 2015 and 2016 have no repeating digits.

Conclusion: In total $2016 - 9 - 81 - 648 - 504 - 4 = 770$ positive numbers from 1 up to 2016 have at least one repeating digit.

The trick is to count the numbers without repeating digits then subtract it from 2016. Clearly, there are 9 1-digit number without repeating digits. There are $2 \binom{10}{2}-9=81$ two-digit numbers without repeatition. There are $3!\binom{10}{3}-2\binom{9}{2}=648$ three-digit numbers without repeatition. If the thousands digit is 1, there are $3!\binom{9}{3}=504$ four-digit numbers without repeatition. When the thousands digit is 2, there are 4 such numbers, namely 2013,2014, 2015, 2016. Hence, there are $504+4=508$ four-digit numbers less than or equal to 2016 without repeating digits. Therefore, the number of integers from 1 to 2016 including 1 and 2016 with repeating digits is given by $2016-(9+81+648+508)=\boxed{770}$