counting

Tricky problem..I do the following way.
Consider all numbers which are between 324,700 and 399999 with the given condition:
First digit: 3 = 1
Second: 2-9 = 8
Third: 4-9 = 6
Fourth: 7-9 = 3
Fifth: 2 = 1
Sixth : 1 = 1
Total possibilities = 1*8*6*3*1*1 = 144

Consider all numbers which are between 400000 and 458600 with the given condition:
First digit: 4 = 1
Second: 0-5 = 6
Third: 0-8 = 9
Fourth: 0-6 = 7
Fifth: 2 = 1
Sixth : 1 = 1
Total possibilities = 1*6*9*7*1*1 = 378

Hence ans = 144+ 378 = 522
What is the answer?

It's because of questions like this that I said above: "It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest - smallest + 1) is normally preferable."

You can't easily apply the multiplication principle from counting to the problem above, without breaking the problem into several cases. The number of choices, for example, for the 'thousands' digit is not 5; it can be anything from 0 to 9, since 330,721 is, for example, a valid number here, as is 339,721. The problem here is that the number of choices you have, say, for the ten thousands' digit depends on what you choose for the hundred thousands' digit. If you choose a '3' for the hundred thousands' digit, you have eight choices for the ten thousands' digit, and if you choose a '4' for the hundred thousands', you only have six choices for the ten thousands' digit. If you really want to use the product rule to count here, you can only resolve this by considering these two cases separately. You'll then discover at each stage you have to consider different cases, and the problem becomes a bit of a mess. It's not a good approach to solving the problem. That said, you've already noticed a good method here, so I'm not sure why you're looking for another one: