Question 1061960: Suppose you are a store manager for a department store and you have 10 floors. you want to put a display ad by the elevators on each floor. 3 displays are for store credit cards,3 displays are for the new store shopping app,2displays are for new cafe and 2 displays are for the store facebook page. How many permutation would you have ??
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 3 displays for store credit cards.
3 displays for new store shopping app.
2 displays for new cafe.
2 displays for for store facebook page.
assuming each of the displays of the same type are different, then you would have 10! different ways to arrange them.
assuming each of the displays of the same type are the same, then you would have 10! / (3! * 3! * 2! * 2!) different ways to arrange them.
lets's do this with less floors and displays and see what we come up with.
assume 3 floors with 2 displays of credit cards and 1 display of new cafe.
if each display is different, then:
the credit card displays are a,b.
the new cafe display is c.
if each display is the same, then:
the credit card displays are a,a.
the new cafe display is b.
3! is equal to 6.
if each display of the same type is different, then the number of permutations is 3! = 6.
they would be:
abc
acb
bac
bca
cab
cba
the first position letter would be on the first floor.
the second position letter would be on the second floor.
the third position letter would be on the third floor.
if each display of the same type is the same, then the number of permutations is 3! / (2! * 1!) = 6 / 2 = 3.
they would be:
aab
aba
baa
the first position letter would be on the first floor.
the second position letter would be on the second floor.
the third position letter would be on the third floor.
if you took the first set of permutations and replaced c with a, you would get:
abc = aba *** 1
acb = aab *** 2
bac = baa *** 3
bca = baa *** 3
cab = aab *** 2
cba = aba *** 1
each pair of unique sets now becomes the same.
abc and cba both become aba
acb and cab both become aab
bac and bca both become baa
so, if each of the displays of the same type are different from each other, you have 10! different ways to display them, and ifr each of the displays fo the same type are the same as each other, you have 10! / (3! * 3! * 2! * 2!) different ways to display them.
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