If there are four bakeries that each close one day a week, how many schedules are possible if at least one bakery is open each day?

Question

I’m self studying for a probabilities and statistic exam so unfortunately, i don’t have anyone to ask. So the question goes, we have 4 bakeries and 7 days in a week. Each bakery closes once a week.

a) How many possibilities are there? Several bakeries may close on the same day.

I answered this with: $7 \cdot 7 \cdot 7 \cdot 7= 2401$.

b) at most one bakery can close on any day.

I answered with: $7 \cdot 6 \cdot 5 \cdot 4$.

c) at least one bakery needs to be open on each day.

Here I am struggling, so any help is much much appreciated, thank you!

Answer 1

• We need the no of possible schedules where each bakery closes on any one day of the week and atleast one bakery is open every day.
• In part A, you found all the schedules possible, but that number contained the days when all bakery are closed.
• Note-(number of schedules when atleast one bakery is open)+(number of schedules when all bakeries are closed one same day)=(All possible Schedules).
• Also, as there are 7 days in the week, possible no. of schedules when all bakeries are closed on same day=7.
• Thus, answer = $7^4 - 7$