First we will do the problem as if we could tell the difference between
the three 4's and the two 6's.

Then, we could choose the 1st digit 6 ways.  (We cannot use 0)
Then, we could choose the 2nd digit 6 ways.
Then, we could choose the 3rd digit 5 ways.
Then, we could choose the 4th digit 4 ways.
Then, we could choose the 5th digit 3 ways.
Then, we could choose the 6th digit 2 ways.
Then, we could choose the 7th digit only 1 way.

That would be 6·6·5·4·3·2·1 = 4320 ways.

If the three 4's and the two 6's were distinguishable, for instance, if
the like digits were different colors, then 4320 would be the final 
answer.  However, they're not distinguishable, so let's look at a random 
sample permutation, say, this one:

6045464.

The answer 6·6·5·4·3·2·1 = 4320 counts all 12 of the following
arrangements separately:

6045464
6045464
6045464
6045464
6045464
6045464
6045464
6045464
6045464
6045464
6045464
6045464

But we cannot tell them apart.  So 6·6·5·4·3·2·1 = 4320 counte EVERY
arrangement 12 times too many, since we cannot tell any of those 12 from
6045464.  So we must divide 4320 by 12, and get the answer 360.  

But how did we know to divide by 12?  That's because there are 3! ways
to place the 3 different colored 3's in the above colored list and for 
each of those 3 different colored 3's, there are 2! ways to place the
colored 2's.  So we divide by 3!2! = 6×2 = 12 

So the answer is

 =  = 360.

Edwin