Answer:
a) 560 (as you know)
Step-by-step explanation:
There are 7C3 = 35 ways to choose 3 full-time days from 7.
There are 2^4 = 16 ways to choose a half-day each day from the remaining 4 days.
The total number of possible schedules is ...
35×16 = 560
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nCk = n!/(k!(n-k)!)
7C3 = 7·6·5/(3·2·1) = 7·5 = 35