How many 8 digit numbers can be formed with exactly 2 pairs from the digits 0-9 where 0 can go in the first place? I'm working on the problem "how many $8$ digit numbers can be formed with the digits $0-9$ where $0$ can also come in the first place with exactly $2$ identical pairs "
So I have $2$ ways to do it, and $1$ is wrong and I'm struggling to understand how to fix it if possible at all.
First way: Each number satisfying the condition of the problem will have $6$ unique digits from the set 0,... 10 and those digits can be chosen in $\binom{10}{6}$ ways. Out of those $6$ digits the $2$ that will form the pair can be chosen in $\binom{6}{2}$ ways. Finally the number of permutations of $8$ characters with $2$ pairs is $\frac{8!}{2! 2!}$ Multiplying those yields the answer.
2nd way: The number of permutations of a sequence AABBCEDF is $\frac{8!}{2! 2!}$. Now A can be choosen in 10 ways, B in 9 ways, E in 8 ways etc to yield the number $\frac{8!}{2! 2!}*10*9*8*7*6*5$ I understand how I'm double counting here but I want to know if there is an approach to this problem along this line of thought. Can I resolve the double counting? This answer comes out $48$ times larger than the correct first answer i.e in my calculation im missing a factor of $\frac{1}{48}$.
 A: You do not need to make an ordered selection of the two digits that will appear twice or an ordered selection of the digits that will appear once since you have accounted for the order in which the digits appear in string by counting the 
$$\frac{8!}{2!2!}$$
permutations of the sequence AABBCDEF.  This is what JMoravitz was hinting at in the comments.
We can choose the two digits that each appear twice in the string in $\binom{10}{2}$ ways.  We can choose the four digits that each appear once in the string from the eight remaining digits in $\binom{8}{4}$ ways.  Therefore, the number of eight-digit decimal strings in which exactly two digits appear twice and the rest of the digits appear once is 
$$\binom{10}{2}\binom{8}{4}\frac{8!}{2!2!}$$ 
Notice that 
$$10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = \binom{10}{2}\binom{8}{4} \cdot 2!4!$$
since 
$$\binom{10}{2}\binom{8}{4} \cdot 2!4! = \frac{10!}{2!8!} \cdot \frac{8!}{4!4!} \cdot 2!4! = \frac{10!}{4!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$$
By choosing the two numbers that appear twice in the string in order, you introduced an extra factor of $2!$; by choosing the four numbers that appear once in the string in order, you introduced an extra factor of $4!$.  This is why your second answer was $2!4! = 48$ times your correct first answer.
