# How many possible different weekly schedules are there if an employee works five full days and two half-days?

A firm works $$7$$ days a week. Every employee must work exactly $$5$$ full days and $$2$$ half-days each week. A half-day can be either morning or afternoon, and two half-days cannot be held on the same day. How many possible different weekly schedules are there?

I have tried $$5{7\choose5}+2{7\choose2}$$ but i am still getting the incorrect answer. The correct answer is $$84$$.

Can someone explain what I am doing wrong in my working out? Thanks

• Once you have chosen which half-days to work, there is no choice for the full days. – saulspatz Aug 7 at 21:24
• It’s worth noting how in the two answers you’ve gotten so far, you can either choose which days are to be the half-days or which days are to be full work days, and you will end up with the same result. – amd Aug 7 at 22:55

The error you have made is in choosing the days to work half days and choosing the days to work full days separately.

Out of the $$7$$ days, the employee must work every day. Two of those will be half days. This gives us an answer of $$2^2\cdot\binom{7}{2} = 84$$

The employee must choose which two days to work half days, and whether to work in the morning or afternoon.

• no. of ways to choose the $$5$$ full days is $$\binom{7}{5}$$.
• Once those are chosen, the two half days are also simultaneously chosen. So you cannot make any independent choices for them anymore. Think of it as follows: if we choose M, Tu, We, Fr, Su as full days then automatically Thu, Sat will have to be half days.
• Now on those two half-days, one can either work in the morning or in the evening. So there are $$4$$ ways to do this $$MM, ME, EM, EE$$.

So the total no. of ways is $$4 \cdot \binom{7}{5}=84.$$

First we will select 5 Full days in 7C5 = 21 ways. Now on to half days each day can be worked in 2 ways =2x2 = 4 Hence for a week it will be 21x4=84 ways.