A problem on combinations

The numbers we're interested in begin with one of $3,4,5,6,7$ (we trivially discount $8000$ ). They end in an odd digit, one of $1,3,5,7,9$ .

If the first digit is odd , there are $4$ choices for the final digit; then $8$ choices for the second and $7$ for the third digit, for a total of $4\times 8\times 7=224$ numbers.

If the first digit is even , there are $5$ choices for the final digit; then $8$ choices for the second and $7$ for the third digit, for a total of $5\times 8\times 7=280$ numbers.

Three of the possible first digits are odd and two are even; so there are a total of $3\times 224 + 2\times 280 = \boxed{1232}$ numbers satisfying the conditions.