Permutations - restriction of having number be greater than and even (no repetition of numbers)

Question

Using $6$ numbers - $3,4,5,6,7,9$; a $6$-digit number is to be formed. What is the working to find out how many of these $6$ digit numbers are even and greater than $500,000$?

My attempt:

When the first digit can be either $5,7,9$ and the last digit can be either $4,6$ - the possible amount of numbers is $3 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 2$. (I'm getting this part correct according to the mark scheme). When the first digit can be either $5,6,7,9$ and the last digit can only be $4$, the possible ammount of numbers is $4 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 1$ (I'm getting this part wrong according to the mark scheme). The answer I am getting is $240$. The current answer is $168$.

Answer 1

Observe that the leading digit must be at least $5$. Consider cases, depending on whether the leading digit is even or odd:

1. If the leading digit is odd, the units digit may be selected from $4$ or $6$. Choose the leading digit from among $5$, $7$, or $9$, choose the units digit, then arrange the remaining digits between them.
2. If the leading digit is $6$, the units digit must be $4$. Arrange the remaining digits between them.

In case 1, the number can be selected in $3 \cdot 2 \cdot 4!$ ways. In case 2, it can be selected in $4!$ ways. Hence, there are $$3 \cdot 2 \cdot 4! + 4! = (3 \cdot 2 + 1)4! = 7 \cdot 4! = 168$$ admissible arrangements.