In a service room, five attendants are scheduled to work together. Knowing that two of them cannot work on the same shift, in how many different ways is it possible to organize the five attendants into two shifts so that the two attendants who cannot work together are on different shifts?

1. Let's label the five attendants as \( A, B, C, D, \) and \( E \), where \( A \) and \( B \) are the two attendants who cannot work together.
2. Since \( A \) and \( B \) cannot be in the same shift, they must be in different shifts.
3. We now need to organize the remaining three attendants \( C, D, \) and \( E \) into the two shifts with \( A \) and \( B \).

To do this:
- Choose one of the 3 remaining attendants to be in the same shift as \( A \):
\binom{3}{1} = 3
- The other 2 attendants will necessarily be in the same shift as \( B \).

- Each of these configurations can happen in two different shift arrangements:
2

Therefore, the total number of ways to arrange the shifts is given by the combination of choosing 1 attendant to be with \( A \) and the arrangement of shifts:
3 \times 2 = 6

Finally,
- For the remaining 3 attendants, the distribution can be selected by choosing 2 out of 3 to be on the shift with \( B \), which is:
\binom{3}{2} = 3
- Multiple that by the other arrangements previously calculated: \(2 \)

Thus, the total number of ways to arrange the attendants is:
3 \times 2 \times 5 \times 2 = 60

Answer:
60