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Mary has a closet of 777 types of shirts and 3 types of shoes. Mary never wears exactly the same thing on two consecutive days (therefore, each time, either the shoe or the shirt (or both) is different from what she wore yesterday). In how many ways can she plan her outfit for the next 15 days?

I've been stuck on this problem for a long time and I just don't know what to do or what I'm even doing is correct.

So far i took the combination of the shirts for 15 days:

777C1
776C1
775C1
774C1
773C1
772C1
771C1
770C1
769C1
768C1
767C1
766C1
765C1
764C1
763C1

I'm not sure if this approach is right :/

George
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1 Answers1

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On day 1, she has a total of $777\cdot 3$ options. (total shirts time total shoes) On day 2, she can have $777\cdot 3 -1$ options, because you have to take out the one outfit that she wore on day 1. Similarly, on all days other than day 1, she has $777 \cdot 3 -1$ options....so it's just $(777\cdot 3)(777\cdot 3 -1)^{14}$

Alan
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