How many different 4 digit combinations will include at least one 7, assuming numbers cannot repeat I cannot get the correct answer - $2016$.
What I have tried so far is thus: the number $7$ can occur $1, 2, 3,$ or $4$ times. Since it is a combination we do not care if the number starts with zero or not since it is a combination and not a number. Therefore:
$(1*9*8*7 + 1*1*8*7 + 1*1*1*7 )*4 = 2268$
where $4$ is the order it can appear.
What should i do instead to get it correct?

If there has to be exactly one 7, that 7 can go into 4 positions, and the other 3 digits can be arranged in $9 \cdot 8 \cdot 7$ ways. In all, $4 \cdot 9 \cdot 8 \cdot 6$ ways. If starting with 0 is not allowed, the ways of starting 0 in the above is $3 \cdot 8 \cdot 7$ ways, so you get $4 \cdot 9 \cdot 8 \cdot 6 - 3 \cdot 8 \cdot 7 = 1848$
If the 7 can repeat, you have to select 1, 2, 3, or 4 positions for the 7s; the others have to be arranged into the rest of the positions (here $x^{\underline{n}} = x (x - 1) \cdots (x - n + 1)$, for $n$ factors). By the same reasoning above:
$$
\binom{4}{1} \cdot 9^{\underline{3}} 
  + \binom{4}{2} \cdot 9^{\underline{2}}
  + \binom{4}{3} \cdot 9^{\underline{1}}
  + \binom{4}{4} \cdot 9^{\underline{0}}
  - \binom{3}{1} \cdot 8^{\underline{2}}
  - \binom{3}{2} \cdot 8^{\underline{1}}
  - \binom{3}{3} \cdot 8^{\underline{0}}
  = 1957
$$
It seems the place you've got this from is mistaken.
