SOLUTION: How many ways can the digits 1, 2, 3, 4, and 5 be arranged in a row of 3 to form an odd counting number if repetition is permitted and if 3 cannot be second and 1 cannot be third?

There are no restrictions on the first digit, so we can
choose it any of 5 ways.

For each of the 5 ways to choose the first digit there are
4 ways to choose the 2nd digit, either 1,2,4, or 5. We are
told it can't be 3. That's 5*4 or 20 ways to choose the first 
two digits.

An odd number must end in an odd digit, so the third digit is 
either 1,3, or 5. But we are told 1 cannot be third, so it can 
only be 3 or 5. So for each of the 20 ways to choose the first 
and second digits, there are 2 ways to choose the last digit.

That makes 20*2 or 40 ways total.

Here they all are:

1.  113
2.  115
3.  123
4.  125
5.  143
6.  145
7.  153
8.  155
9.  213
10.  215
11.  223
12.  225
13.  243
14.  245
15.  253
16.  255
17.  313
18.  315
19.  323
20.  325
21.  343
22.  345
23.  353
24.  355
25.  413
26.  415
27.  423
28.  425
29.  443
30.  445
31.  453
32.  455
33.  513
34.  515
35.  523
36.  525
37.  543
38.  545
39.  553
40.  555

Edwin