A question on permutation

Question

A credit card number has 4 digits, and it's owner forgot them.Presuming he knows the following details, how many options does he have to choose from before reaching the right combination:

1. First digit is either a 3 or 1.
2. Second digit differs from the first one.
3. Third digit is greater than 6.
4. fourth digit is greater than the third one.

Attempt at a solution:

There are 2 options for first digit, 9 for second. Third and fourth is a bit complicated, since we know fourth has to be bigger than third, hence third number can be either 7 or 8 (2 options) and hence the fourth 8 or 9 , thus further 2 options for a fourth number.

Summing up: n = 2*9*2*2 = 72.

Answers say 54 though.

Any Help would be appreciated.

Answer 1

There are $2$ options for first number and $9$ for second then if third is $7$ there are two options for fourth and if third is $8$ there is one option for fourth :

so we have $2*9*2*1+2*9*1*1=54$