Combinatorics problem involving selection of digits How many 10-digit decimal sequences using $(0, 1, 2, . . . , 9)$ are there
in which digits 3, 4, 5, 6 all appear?
What I did to solve this question was this. 
The number of ways to select $3,4,5,6$ from $10$ numbers is 
$$\binom{10}{4}$$
and the ways to fill the rest of the digits would be $10^{10-4}=10^{6}$
So I thought that the total number of possibilities would 
$$\binom{10}{4}*10^{6}$$
I was wondering what is wrong with this reasoning? 
The answer that you get using inclusion-exclusion principle is different. 

First, your description of $10 \choose 4$ is wrong because there is only one way to select specifically those digits.  What you really should be saying is there are $10 \choose 4$ ways to select the positions for the $3,4,5,6$.  Second you should multiply by $4!$ for the orders of $3,4,5,6$  Third, you are double counting numbers that have additional copies of $3,4,5,$ and/or $6$.  For example $3456311111$ is counted twice, once when you count the first $3$ in the specified $3,4,5,6$ and again when you count the second $3$
