$4$ digit integers such that none of the digits appear more than twice 
How many $4$ digit integers can be chosen such that none of the digits
  appear more than twice?

$4$-digit integers with distinct digits= $9\times9\times8\times7 = 4536$
$4$-digit integers with two alike digits $(XXYY) = 9\times9\times2\times1\times \cfrac{4!}{2!\times2!} = 972$
Total = $5508$
Is the solution correct?
 A: Direct approach. There are three possibilities.  The first is ABCD, where all four digits are distinct.  There are $9 \times 9 \times 8 \times 7 = 4536$ of these.
The second is AABC/ABAC/ABCA/BAAC/BACA/BCAA.  Each of these has $9$ choices for the first position, and then $9 \times 8$ choices for the other two distinct digits.  There are thus $6 \times 9 \times 9 \times 8 = 3888$ of these.
Finally, we have AABB/ABAB/ABBA.  Each of these has $9$ choices for A and $9$ choices for B, so there are $3 \times 9 \times 9 = 243$ of these.
The total is $4536+3888+243 = 8667$.

If we allow leading zeros, the totals are $5040+4320+270 = 9630$.


Alternative approach. Count up the number of numbers that don't qualify, and then subtract from $9000$ (the number of numbers from $1000$ through $9999$, inclusive).
There are just $9$ numbers that have four of a kind: $1111, 2222, \ldots, 9999$.
There are also $9$ numbers that have three of a kind with three $0$'s: $1000, 2000, \ldots, 9000$.
For numbers that have three of a kind with three of a digit other than $0$: There are $9$ choices.  If the other digit is in the first position, it can be anything other than the chosen digit or $0$, so there are $8$ choices.  Otherwise, there are $9$ choices, for a total of $8+9+9+9 = 35$ possibilities for the other digit and its placement.  This therefore accounts for $9 \times 35 = 315$ further disallowed numbers.
The total number of disallowed numbers is $9+9+315 = 333$, so the number of qualifying numbers must be $8667$.

If we allow leading zeros, the total number of possibilities increases to $10000$, and the number of disallowed numbers is $10+10 \times 4 \times 9 = 370$, so the number of permitted numbers is $9630$.

A: Take the total amount of numbers: $9\cdot10\cdot10\cdot10=9000$.
Subtract the amount of numbers with a digit which occurs exactly $4$ times: $9$.
Subtract the amount of numbers with a digit which occurs exactly $3$ times:


*

*With $0$ occurrences of $0$, there are $9\cdot8\cdot4=288$ such numbers

*With $1$ occurrences of $0$, there are $9\cdot3=27$ such numbers

*With $3$ occurrences of $0$, there are $9$ such numbers



The answer is therefore $9000-9-288-27-9=8667$.
