You can use in to test if a substring appears:

>>> 'code' in 'recoded'
True

This removes the need to loop over the string.

You can generate your test strings by looping over all letters in the alphabet; Python already has the latter available for you at string.ascii_lowercase and a list comprehension:

import string

possibilities = ['co{l}e'.format(l) for l in string.ascii_lowercase]

count = 0
for possibility in possibilities:
    if possibility in inputstring:
        count += 1

You could also just test for co appearing, and see if there is a letter e further along. You can use the str.find() method to find the position of an occurrence and search from there; str.find() takes a starting position to search for the next match:

count = 0
start = 0
while True:
    position = inputstring.find('co')
    if position == -1:
        # not found, end the search
        break
    if len(inputstring) > position + 2 and inputstring[position + 2] == 'e':
        count += 1
    start = position + 1

However, most experienced programmers will use regular expressions to find such matches:

import re

count = len(re.find_all(r'co[a-z]e', inputstring))

Here the expression uses [a-z] to match a single character as a class, anything in that series (so letters from a to z) would match. The re.findall() function returns a list of all matches found in the input string, so all you have to do then is take the len() length of that list to get a count.

You can use in to test if a substring appears:

>>> 'code' in 'recoded'
True

This removes the need to loop over the string.

You can generate your test strings by looping over all letters in the alphabet; Python already has the latter available for you at string.ascii_lowercase and a list comprehension:

import string

possibilities = ['co{l}e'.format(l) for l in string.ascii_lowercase]

count = 0
for possibility in possibilities:
    if possibility in inputstring:
        count += 1

You could also just test for co appearing, and see if there is a letter e further along. You can use the str.find() method to find the position of an occurrence and search from there; str.find() takes a starting position to search for the next match:

count = 0
start = 0
while True:
    position = inputstring.find('co')
    if position == -1:
        # not found, end the search
        break
    if len(inputstring) > position + 2 and inputstring[position + 2] == 'e':
        count += 1
    start = position + 1

However, most experienced programmers will use regular expressions to find such matches:

import re

count = len(re.findall(r'co[a-z]e', inputstring))

Here the expression uses [a-z] to match a single character as a class, anything in that series (so letters from a to z) would match. The re.findall() function returns a list of all matches found in the input string, so all you have to do then is take the len() length of that list to get a count.