How many such numbers are possible

Let $abcde$ be the required integer (where, $a,b,c,d~and~e$ are digits).

It can be seen that: $\begin{array}{c}& a\in \{2,3,4,5,6\} \\ b,c,d\in \{0,1,2,\cdots,9\}\\ e\in \{0,2,4,6,8\}\end{array}$

We can have two disjoint cases:

Case 1: If $a$ is a even digit then, $a$ has $3$ possibilities and $e$ has $4$ possibilities and the remaining digits have $\binom{8}{3}\cdot 3!$ possibilities. So, total number of such even numbers possible are $\displaystyle 3\times 4\times \binom{8}{3}\cdot 3!=4032$

Case 2: If $a$ is a odd digit then, $a$ has $2$ possibilities and $e$ has $5$ possibilities and the remaining digits have $\binom{8}{3}\cdot 3!$ possibilities. So, total number of such even numbers possible are $\displaystyle 2\times 5\times \binom{8}{3}\cdot 3!=3360$

Thus, the total number of even numbers between $20000~and~70000$ are $4032+3360=\boxed{7392}$ in number.

If the first digit is $2,4,6$ ,then the last digit has $4$ possibilities,

If the first digit is $3,5$ ,then the last digit has $5$ possibilities.

The other three digits has $8\times7\times6$ possibilities.

Hence,there are $(4+5+4+5+4)\times8\times7\times6=7392$ possibilities.