<s>[INST] Permutations - restriction of having number be greater than and even (no repetition of numbers) 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 $168$.
 [/INST] Observe that the leading digit must be at least $5$.  Consider cases, depending on whether the leading digit is even or odd:  


*

*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.

*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. 

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