I'm not sure what the phrase "no two or three athletes finish at the same time" is doing in the question - if no two athletes finish at the same time, clearly no three athletes finish at the same time.

Anyway, with no restrictions, there are 5! = 120 orders in which they could finish. If you think just of arranging A, B and C, there are six ways to do that, and in only one of those six arrangements is it true that A is ahead of B and B ahead of C. So if you pick a random arrangement of A, B, C, D and E, there's a 1/6 chance that A is ahead of B who is ahead of C, or in other words, the fraction of all arrangements that meet that condition is 1/6. So the answer is 1/6 * 120 = 20.

Hi NandishSS,

With 5 runners who finish - and no "ties", there are 5! = 120 possible outcomes. We certainly don't want to list all of those out, so we have to narrow down the possibilities. We can do that in a number of different ways; here's a way that's more 'visual' in nature:

We're looking for the outcomes in which A finishes before B and B finishes before C. For example...
ABC _ _
With this option, D and E would take up the missing two spots (as either DE or ED), so this order of ABC in the first 3 'spots' represents 2 options that fit what we're looking for. If we could map out all of the other options, then we'd have the total....

ABC _ _
AB _ C _
AB _ _ C
A _ BC _
A _ B _ C
A_ _ BC
_ ABC _
_ AB _ C
_ A _ BC
_ _ ABC

Each of these 10 orders includes 2 possible outcomes that fit what we're looking for. So the final answer is (10)(2) = 20

Final Answer:
B

GMAT assassins aren't born, they're made,
Rich

NandishSS wrote: ↑

Thu Jul 20, 2017 5:07 am

Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

OA: B

Source : MathRevolution

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