Possibilities with unit digits and numbers

Question

$x$ is a three digit number greater than $700$. If $x$ is an odd number and each digit is not equal to zero, what is the possible number of $x$? (Replacement is not allowed)

Answer: $91$

Can somebody please explain how to solve this problem? I did $3\cdot9\cdot5 = 135$ possibilities, but since replacements are now allowed I tried $3\cdot8\cdot4$ but still does not give me an answer.

Answer 1

We will have to make three different cases with 7, 8, and 9 at Hundred's place

1. "7" at Hundred's place -> 7 at tens (2,4,6,8 and any 3 of the remaining odds) and 4 at ones (1,3,5, ,9) total , $7\cdot4=28$
2. "8" at hundred -> 7 at tens (2,4,6 and any 4 of the remaining odds) and 5 at ones (1,3,5,7,9) , $7\cdot5=35$
3. "9" at Hundred's place -> 7 at tens (2,4,6,8 and any 3 of the remaining odds) and 4 at ones (1,3,5,7, ) total , $7\cdot4=28$

Total numbers, $28+28+35=91.$