Work-at-home days such that the office is always staffed I am quite rusty on combinatorial formulas, but I think the following practical question has a combinatorial answer.  This is not homework.
A company gives it's technicians two work days per week to work at home.  In a team of four, a manager and three technicians, what assignment of days working at home will gurantee that at least one technician or the manager is NOT at home on every working day (Mon-Fri)?  I.E., at least one person from the team of four is in the office every working day.
There are only 10 unique combinations of two weekdays (5 things taken 2 at a time), so I think the question becomes which of the combinations of two days among four employees satisfies the constraint that the count of any one of the days assigned to work at home among the four employees is less than 4 (thus guaranteeing that for any weekday at least one employee is NOT at home)?
If I haven't formulated this question correctly, please enlighten me and help cure my ignorance (and rusty mathematics).
Regards,
Peter
[Edit] Based on one of the answers I have seen so far, I need to clarify two things:


*

*All four team members always have two scheduled days at home each week

*I would like to know if there is a way to generate combinations of scheduled days at home for the whole team which satisfy the constraint of at least one team member not at home each weekday

 A: The team has a total of eight days off and twelve days in per week.  You can get that with one person at home two of the days and two people at home the other three days.  You have two or three people in the office each day.
There are many other combinations, but it is easy to guarantee at least one person in the office or even two.
A: This is an inclusion-exclusion problem.
Call $S$ the set of all $10^4=10,000$ schedules.  Let $M$ denote the subset wherein all four take a day off on Monday.  We calculate $|M|=4^4=256$ -- each of the four people chooses one of four remaining days for their second day off.  Let $T$ denote the subset wherein they all take Tuesday off.  We have $|T|=256$.  Similarly, we have $W,R,F$ for Wednesday, Thursday, Friday.
However, the answer isn't simply $|S|-|M|-|T|-|W|-|R|-|F|=10,000-5*256=8720$.  Consider the schedule where everybody takes Monday and Tuesday off.  We counted it once in $|S|$, but subtracted it twice -- once in $|M|$ and again in $|T|$.  Hence we need to add it back in, so we count it a total of zero times.  There are ${5 \choose 2}=10$ such issues, one for each pair of days.  Thus the answer is $8730$.
