The question at hand is asking for four-digit even numbers with no repeating digits within a certain bounds, such as $1000 \le N < 6000$.
$$N = (n_1)(n_2)(n_3)(n_4)$$
$$n_1 = \{1,2,3,4,5\}$$
$$n_2 = \{0,1,...,9\}$$
$$n_3 = \{0,1,...,9\}$$
$$n_4 = \{0,2,4,6,8\}$$
Suppose $n_4=0$, then the sum is $5 \cdot 8 \cdot 7 \cdot 5$.
Suppose $n_4=2$, then $4 \cdot 8 \cdot 7 \cdot 5$
so the "value" of $n_1$ varies with $n_4=\{2,4\}$...
How do I take this variation into account when producing the final answer?
Thank you for your help and time.
There $5$ options for $n_4$. Two of the choices are $2,4$ and three of them are $0,6,8$.
$n_1$ may be $1,2,3,4$ or $5$. If choices $2$ or $4$ for $n_4$ there are $4$ remaining options for $n_1$. So there are $2*4$ options for $n_1, n_4$ if $n_4 = 2,4$. Otherwise there are $5$ options for $n_1$ if $n_4= 0,6,8$. So there are $3*5$ options for $n_1, n_4$ if $n_4 =0,6,8$.
So there are $2*4 + 3*5$ options for $n_1$ and $n_4$.
$n_2$ may be any of the $10$ digits other than $n_1, n_4$ so there are 8 options. And $n_3$ may be any other than $n_1,n_4,n_2$ so there are 7 options.
So there are $(2*4 + 3*5)*8*7$ such numbers.