How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$? How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$?
My try:
All possible $4$-digit numbers $= 7 \cdot 7 \cdot 7 \cdot 6$ 
$4$-digits numbers NOT containing $3, 5 = 5 \cdot 5 \cdot 5 \cdot 4$
Answer= $7 \cdot 7 \cdot 7 \cdot 6-5 \cdot 5 \cdot 5 \cdot 4 =1558$ 
Is that OK ?  If not, please explain my mistake. 
Thank you. 
 A: You also need to subtract out the number of numbers that contain one or more $5$s but no $3$, as well as the number of numbers that contain one or more $3$s but no $5$.
Or, you can also just count directly:


*

*One $3$, one $5$, two something else

*One $3$, two $5$s, one something else

*Two $3$s, one $5$, one something else

*Three $3$s, one $5$

*Two $3$s, two $5$s

*One $3$, three $5$s

A: Your answer is incorrect, because the negation of the statement that the positive integers includes the digits $3$ and $5$ is that it does not contain the digit $3$ or does not contain the digit $5$.  What you subtracted is the number of $4$-digit positive integers that contain neither the digit $3$ nor the digit $5$.  However, numbers such as $3460$ and $2145$ are also inadmissible. 

How many $4$-digit positive integers can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ with repetition contain the digits $3$ and $5$?

If there were no restrictions, the thousands place could be filled in six ways (since $0$ is excluded) and each of the three remaining places could be filled in seven ways (assuming repetition of digits is permitted).  Hence, there are a total of $$6 \cdot 7 \cdot 7 \cdot 7$$ four-digit positive integers that can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ with repetition.  
From these, we must subtract those that do not contain the digits $3$ and $5$.  Numbers that do not contain the digits $3$ and $5$ do not contain the digit $3$ or do not contain the digit $5$.
How many four-digit postive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition do not contain the digit $3$?
There are five ways to fill the thousands place (since neither $0$ nor $3$ is permitted) and six ways to fill each of the remaining places (since $3$ is not permitted).  Hence, there are $$5 \cdot 6 \cdot 6 \cdot 6$$ such numbers.
By symmetry, there are also 
$$5 \cdot 6 \cdot 6 \cdot 6$$
four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition that do not contain the digit $5$.
However, if we subtract those numbers that do not contain the digit $3$ and those numbers that do not contain the digit $5$ from the total, we will have subtracted those numbers that contain neither the digit $3$ nor the digit $5$ twice.  We only want to subtract them once, so we must add them to the total.
How many four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition contain neither the digit $3$ nor the digit $5$?
There are four ways to fill the thousands place (since neither $0$ nor $3$ nor $5$ is permitted) and five ways to fill each of the remaining three places (since neither $3$ nor $5$ is permitted).  Hence, there are 
$$4 \cdot 5 \cdot 5 \cdot 5$$ such numbers.
By the Inclusion-Exclusion Principle, the number of positive four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition that contain the digits $3$ and $5$ is 
$$6 \cdot 7 \cdot 7 \cdot 7 - 2 \cdot 5 \cdot 6 \cdot 6 \cdot + 4 \cdot 5 \cdot 5 \cdot 5$$ 
More formally, let's follow the suggestion of Nicolas FRANCOIS made in the comments.  Let the universal set, $U$, be the set of four-digit positive integers that may be formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition; let $A$ be the event that such a four-digit positive integer includes the digit $3$; let $B$ be the event that such a four-digit positive integer includes the digit $5$.  What we wish to calculate is $$|A \cap B| = |U| - |(A \cap B)^C| = |U| - |A^C \cup B^C| = |U| - (|A^C| + |B^C| - |A^C \cap B^C|)$$
What we showed above is that 
\begin{align*}
|U| & = 6 \cdot 7 \cdot 7 \cdot 7\\
|A^C| & = 5 \cdot 6 \cdot 6 \cdot 6\\
|B^C| & = 5 \cdot 6 \cdot 6 \cdot 6\\
|A^C \cap B^C| & = 4 \cdot 5 \cdot 5 \cdot 5
\end{align*}
which gives 
\begin{align*}
|A \cap B| & = |U| - |(A \cap B)^C|\\
           & = |U| - |A^C \cup B^C|\\
           & = |U| - (|A^C| + |B^C| - |A^C \cap B^C|)\\
           & = |U| - |A^C| - |B^C| + |A^C \cap B^C|\\
           & = 6 \cdot 7 \cdot 7 \cdot 7 - 5 \cdot 6 \cdot 6 \cdot 6 - 5 \cdot 6 \cdot 6 \cdot 6 + 4 \cdot 5 \cdot 5 \cdot 5\\
           & = 6 \cdot 7 \cdot 7 \cdot 7 - 2 \cdot 5 \cdot 6 \cdot 6\cdot 6 + 4 \cdot 5 \cdot 5 \cdot 5
\end{align*}

How many $4$-digit positive integers can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ without repetition contain the digits $3$ and $5$?

In this problem, either $3$ or $5$ is the leading digit or neither is.
$3$ or $5$ is the leading digit:


*

*Choose which of them is the leading digit.

*Choose the place occupied by whichever number in the set $\{3, 5\}$ that has not been used as the leading digit.  

*Now that $3$ and $5$ have been placed, choose which of the remaining numbers occupies the first open place.

*Choose which of the remaining numbers fills the final open place.



 There are $$2 \cdot 3 \cdot 5 \cdot 4$$ such numbers.

Neither $3$ nor $5$ is the leading digit:


*

*Choose which of the other non-zero digits occupies the thousands place.

*Choose which of the remaining places is occupied by the digit $3$.

*Choose which of the remaining places is occupied by the digit $5$.

*Choose which of the remaining digits fills the remaining place.



  There are $$4 \cdot 3 \cdot 2 \cdot 4$$ such numbers.

Since these cases are mutually exclusive and exhaustive, the answer is found by adding the results for the two cases.
A: Since you only care about 4 states: contains neither, contains 3 not 5, contains 5 not 3, contains both; you can use a transition matrix:
$$\begin{bmatrix} 4 & 1 & 1 & 0 \end{bmatrix}
\begin{bmatrix} 
5 & 1 & 1 & 0 \\
0 & 6 & 0 & 1 \\
0 & 0 & 6 & 1 \\
0 & 0 & 0 & 7 \end{bmatrix}^3
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$
This way, if you wanted to change it to "how many 55 digit numbers..."  all you have to do is change the $3$ in the exponent to a $54$.
Interesting that if you diagonalize the matrix you the same expression as N.G.Taussig's answer: $6 \times 7^3 - 2 \times 5 \times 6^3 + 4 \times 5^3$
