Problema Solution
Ten different coloured pennants are hung in a row to decorate a gymnasium wall. In how many different orders can the pennants be hung?
Answer provided by our tutors
The number of colors that could go into the first spot is 10, then there are 9 left for the second spot, and 8 for the third spot and so on.. so we have
10*9*8*7*6*5*4*3*2*1 = 10! = 3,628,800 different orders
or we can easily conclude that this is a permutation without repetition Pn = n! where n = 10 so P10 = 10! = 3,628,800
the pennants can be hung in 3,628,800 different orders.