looking for 8 digit numbers with 4 digits being used twice I'm looking for $8$ digit numbers with $4$ digits being used twice. for example : $11223344$ and $12123434$ and $11002233$
Its not allowed to use one digit like: $11223345$
for $4$ digit numbers with $2$ digits being used twice I have computed $243$ numbers. 
I need to find out how many 8-digit numbers satisfy afore-mentioned conditon
 A: The number of ways to pair up the digit positions is $7 \cdot 5 \cdot 3 \cdot 1 = 105$.  For each such pairing, to get a number you assign the first pair (including the leading digit) any of the $9$ possible digits $1,2,\ldots,9$ (since we don't want a leading $0$), the next pair any of $9$ remaining possibilities (including $0$ this time, but not the one assigned to the first pair), the next any of $8$, and the last pair any of $7$.  Thus there are
$105 \cdot 9 \cdot 9 \cdot 8 \cdot 7 = 476280$ possible numbers.
A: The first digit on the left can not be $0$.  So there are $9$ possible digits the first digit can be.  Call that first digit $a$.
You will need another $a$ somewhere.  There are $7$ places to put the second $a$.
Now you need a digit to go in the first free space from the left. The digit can be anything except $a$.  Ao there are $9$ possible digits for the digit in the first free space to be.  Call that digit $b$.
You will need another $b$ somewhere.  There are $5$ places to put the second $b$.
Now you need a digit to go in the first free space from the left.  There are $8$ possible digits.  Call it $c$.
Now you need to place a second $c$ somewhere.  There are $3$ places left.
Now at this point there are only two spaces left. They must both have the same digit.  There are $7$ choices for that last digit.
So there  are $9*7*9*5*8*3*7$ possible numbers.
I was very strict on order because I wanted to avoid double counting or any potential taking casese into accont.
A: If $0$ is not included, then we have $9\choose 4$ ways to choose the pairs, and $\frac{8!}{(2!)^4}$ ways to arrange them. If $0$ is included, then there are $9\choose 3$ choices for the other three pairs. Now, the two $0's$ can be arranged in $7\choose 2$ ways, and for the remaining digits we have $\frac{6!}{(2!)^3}$ ways of permuting them. This gives the total ways: $${9\choose 4} \times \frac{8!}{(2!)^4} + {9\choose 3}\times {7\choose 2}\times \frac{6!}{(2!)^3}=\boxed{476280}$$
A: Case 1: $0$ is not included.
Out of nine numbers i.e (1,2,..., 9), selecting any four numbers by $$9\choose4$$ ways and rearranging them in $$\frac{8!}{(2!)^4}$$
$\therefore$ Total numbers=$${9\choose4}{\frac{8!}{(2!)^4}}=317520$$
Case 2: $0$ is included.
Now, two places are reserved for $0$. So we want three numbers to fill six places. That can be done in $9\choose3$ ways. Also $0$ can't occur at first place ($\because$ it will be then $7$ digit number).
They can be arranged in $$\frac{8!}{(2!)^4}-\frac{7!}{(2!)^3}$$ ways.
$\therefore$ Total numbers=$${9\choose3}\left(\frac{8!}{(2!)^4}-\frac{7!}{(2!)^3}\right)=158760$$
Now add the numbers of both the cases to get the final answer as $476280$.
