An eccentric millionaire has 5 golden hooks from which to hang her expensive artwork. She wants to have enough paintings so she can change the order of the arrangement each day for the next 41 years. (The same five paintings are okay as long as the hanging order is different.) What is the fewest number of paintings she can buy and still have a different arrangement every day for the next 41 years?
365 days * 41 years + 10 leap year days = 14,975 days
what is the lowest permutations count of n such that nP5 >= 14,975
We see that 9P5 = 15,120, so the answer is 9 paintings
365 days * 41 years + 10 leap year days = 14,975 days
what is the lowest permutations count of n such that nP5 >= 14,975
We see that 9P5 = 15,120, so the answer is 9 paintings