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?
 A: Hint: this is a trick question. How many times at most can the number 7 occur assuming numbers cannot repeat?
I'm leaving my earlier hint below, because it is a very simple, general method that gets also the right answer in this case.

Hint: think about


*

*how many 4 digit numbers there are altogether where digits don't repeat? 

*how many 4 digit numbers are there with no 7s where digits don't repeat?

How many different 4 digit combinations will include at least one 7, assuming numbers cannot repeat?

How can you define this in terms of 1. and 2.?
A: 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.
A: Answer:  I am assuming that 0 in the first place is possible.  Seems like a password!!
How many 4 digit combination with one 7 will get you the result.  But your question asks for atleast one.  
_ _ _ _ Assume this four digit number and fix 7 in any one place.  This you can do it in four ways - 4.  The second digit from {0-9} except 7 can be chosen in 9 ways.  The third digit except 7 and the digit chosen earlier, can be chosen in 8 ways.  Similarly, the fourth digit can be chosen in 7 ways.  Thus for one 7 to appear in the four digit combination, you can have a total of 4*9*8*7 = 2016 ways.
If you need an answer for 2 7's , 3 7's and 4 7's.  Fix the two 7's and this can be done in 4C2 ways= 6.  The third digit can be chosen in 9 ways, the four digit can be chosen in 8 ways to a total of 432.
Fix the three 7's  and this can be done in 4 ways.  The fourth digit can be chosen in 9 ways to a total of 36.  And finally, four 7 will get you 1.
Thus for atleast one 7, the number of ways = 2016+432+36+1 = 2485.
Good Luck 
Thanks
Satish
A: 2016 = 9 * 8 * 7 * 4.  So, working backwards:
There are 4 different ways of placing the solitary 7 in one of the 4 slots.  After that, there are 9 * 8 * 7 ways of the filling the 3 remaining slots with 3 digits, without repeats, from the remaining 9 possible digits.  Hence the answer.
