Answer to Question #247219 in Statistics and Probability for anonymous456

Question

Suppose that we want to list A, B, C, D, E, F into a sequence in a way

that A and B are placed on the same side of C. For example, A, D, B, E, F, C and F, E, D, C, B, A

would be both desirable sequences because A, D, B, E, F, C has A and B on the left side of C and

F, E, D, C, B, A has A and B on the right side of C. However, A, E, C, F, B, D would not be a

desirable sequence. Note that A B and C do not have to be adjacent to each other. How many

different sequences can you make?

Answer 1

There are 6 letters: A, B, C, D, E, F.

Let us choose places for 3 letters: A, B, and C. We can do it in ${6 \choose 3}=\tfrac{6!}{3!\cdot 3!}=20$ ways.

Other 3 places are for letters D, E, and F.

We can write down letters A, B, and C in 4 ways: A,B,C; B,A,C; C,A,B; C,B,A.

And letters D, E, and F in 3!=6 ways: D,E,F; D,F,E; E,DF; E,F,D; F,D,E; F,E,D.

So, we can make $20\cdot 4\cdot6=480$ different sequences.

Answer: 480.