Step 1: Write each person's one-day work rate.
A finishes the job in 36 days, so A does \(\frac{1}{36}\) of the work in one day.
B finishes it in 54 days, so B does \(\frac{1}{54}\) of the work in one day.
C finishes it in 72 days, so C does \(\frac{1}{72}\) of the work in one day.
Step 2: Let T be the total number of days till the work is finished.
All three start together on day 1.
A works until 8 days before the finish, so A works for \((T-8)\) days.
B works until 12 days before the finish, so B works for \((T-12)\) days.
Since only A and B are said to have left early, C stays till the very end, so C works for the full \(T\) days. This \(T\) is exactly what the question is asking us to find.
Step 3: Write the total work done as 1 complete job.
\[ \frac{T-8}{36} + \frac{T-12}{54} + \frac{T}{72} = 1 \]
Step 4: Clear the fractions.
The LCM of 36, 54 and 72 is 216. Multiply every term by 216:
\[ 6(T-8) + 4(T-12) + 3T = 216 \]
\[ 6T - 48 + 4T - 48 + 3T = 216 \]
\[ 13T - 96 = 216 \]
\[ 13T = 312 \]
\[ T = 24 \]
Step 5: Confirm what the question asks.
Since C worked from day 1 right up to the day the work finished, the number of days C worked equals \(T\), which is 24 days.
Step 6: Check the other options.
Option (1), 48 days, would mean the whole job took twice as long as the equation shows, which does not satisfy the work equation above.
Option (3), 12 days, is too short, since A alone stops 8 days before the finish; if \(T\) were 12, A would only work for 4 days, and the equation would not balance.
Option (4), "None," is not needed since option (2) fits exactly.
Final Answer:
C worked for the entire duration of the job, which is 24 days.
\[ \boxed{24 \text{ days}} \]