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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

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    $\begingroup$ Once you have chosen which half-days to work, there is no choice for the full days. $\endgroup$ – saulspatz Aug 7 at 21:24
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    $\begingroup$ 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. $\endgroup$ – amd Aug 7 at 22:55
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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.

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  • 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 $4 \cdot \binom{7}{5}=84.$

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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.

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