Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each

Solution:

If A, B, and C are all together and in that order, we can treat [ABC] as a single entity, and we have EF[ABC], FE[ABC], [ABC]EF, [ABC]FE, E[ABC]F, or F[ABC]E. We see that there are 6 such ways.

If A, B, and C are separate, we can place one of E and F between one of the two pairs {A, B} and {B, C} and the other between the other pair. That is, the order of the 5 people could be AEBFC or AFBEC. We see that there are 2 such ways.

If A and B are together and C is separate from A and B, we can place one of E and F between AB and C and the other before A or after C OR both E and F between AB and C. That is, the order of the 5 people could be ABECF, FABEC, ABFCE, EABFC, ABEFC, or ABFEC. We see that there are 6 such ways.

There should also be 6 ways if B and C are together and A is separate from B and C. Therefore, there are a total of 6 + 2 + 6 + 6 = 20 ways.

Alternate Solution:

Let’s suppose that A finished before B and B finished before C. Thus, not considering D or E, the arrangement looks like the following:

_ A _ B _ C _

We see that there are four possible positions for the athlete D. After we place D (say to the first position), the arrangement will look like the following:

_ D _ A _ B _ C _

Now, we see that there are five possible positions for athlete E. Notice that it does not matter where we placed D in the previous step; in addition to the four options that were available for D, E can also be placed right before or right after D, thus the number of options will increase by one. Notice also that any ordering where A is before B and B is before C can be obtained this way by placing D and E in appropriate positions.

Thus, there are 4 x 5 = 20 different orderings where A is ahead of B and B is ahead of C.

Answer: B