Combinatorial question: permutation, binomial coefficient

[blob84]

How many numbers of 6 digits which have exatctly the digit 1 (2 times), digit 2 (2 times), without zero, are there?
The book post this solution: \frac{6!}{2!2!}*\binom{7}{2} + \frac{6!}{2!2!2!}*7= 4410,
but I'm trying to find an explanation for this result.

 

[mfb]

That is a weird way to calculate the number.

Here is what I would do:

There are $6 \choose 2$ ways to place the "1"s, and $4 \choose 2$ ways to place the "2"s afterwards. If you write both with factorials, a 4! cancels and you get $\frac{6!}{2! 2! 2!}$. The remaining two digits have 7 options each (3...9), therefore the total number of digits is...

Of course, you can split 7*7 in $2 {7\choose 2} + 7$, but where is the point? The first part corresponds to the number of numbers where the two remaining digits are different, but I don't see a reason to consider them separately.

 

[blob84]

It is \binom{6}{2}*\binom{4}{2}*7^2 by the general form of the product rule, where there are for any pairs of digits 3 cases, if I'm not wrong, thank you.
I suppose to get this weird formula it calculate permutation with repetition and it brokes the problem in two sets.

 

[ssd]

The 1st term corresponds to the case when the other two digits are different and the 2nd term corresponds to the case when another digit is repeated (three double digits).