This question is from McGraw-Hill Ryerson Mathematics of Data Management grade 12 textbook, chapter 4.2, page 245, question 9:
Roberta is a pilot for a small airline. If she flies to Sudbury three times, Timmins twice, and Thunder Bay five times before returning home, how many different itineraries could she follow? Explain you reasoning.
Sudbury --- S, Timmins ---- T, Thunder Bay ---- B
I interpret this to have SSSTTBBBBB arranged in different ways.
This falls into the category of arranging "like" elements
number of arrangements = 10!/(3!2!5!) = 2520
BUT .................
On second thought, this is not as simple as my solution above suggests.
e.g. an itinary of SSTTBSBBBB, which is one of the above arrangements,
wouldn't make any physical sense, that is, travelling from S to S is silly.
We cannot have 2 like destinations side by side.
My new thinking goes like this:
we could have B_B_B_B_B_ or _B_B_B_B_B
in each case that will leave 5 spaces between the B's, with no 2 like
destinations side by side. If we start with something like STB we would have
a BB somewhere further down the line.
for B_B_B_B_B_ , we would have B*5*B*4*B*3*B*2*B*1 ----> 120
but we have 3 S and 2 T, so the number of cases for the above
is 120/(3!2!) = 10
The same would be true for the _B_B_B_B_B case, so
there would be only 20 such itineraries.