Example 1.8 Let the number of 4-digit even numbers with distinct digits formed by $1,2,3,4,5,6$ be $N$, and their sum be denoted as $M$. Find $N$ and $M$.
Solution: For 4-digit even numbers formed by $1,2,3,4,5,6$ with distinct digits and with the unit digit being 2, 4, or 6, there are $P(5,3)=60$ such numbers for each case. Therefore, by the principle of addition, we have
$$
N=60 \times 3=180 \text {. }
$$

Let $a_{1}, a_{2}, a_{3}, a_{4}$ represent the sum of the unit, tens, hundreds, and thousands digits of these 180 even numbers, respectively. Then
$$
M=a_{1}+10 a_{2}+100 a_{3}+1000 a_{4} .
$$

Since among these 180 even numbers, the unit digit is 2, 4, or 6 in 60 numbers each, we have
$$
a_{1}=(2+4+6) \times 60=720 \text {. }
$$

Since among these 180 even numbers, the tens (hundreds, thousands) digit is 1, 3, or 5 in $3 \times P(4,2)=3 \times 4 \times 3=36$ numbers each, and 2, 4, or 6 in $2 \times P(4,2)=2 \times 4 \times 3=24$ numbers each, we have
$$
\begin{aligned}
a_{2} & =a_{3}=a_{4} \\
& =(1+3+5) \times 36+(2+4+6) \times 24 \\
& =612 .
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
M & =720+612 \times(10+100+1000) \\
& =680040 .
\end{aligned}
$$