Short answer:
There are 5!, or 120 possible arrangements. Half the time Meg will beat Bob (and the other half of the time Bob will beat Meg), so Meg is ahead of Bob in 5!/2, or 60, of the arrangements. This is the way the GMAC "wants" you to solve the problem: avoid the clunky casework!
Long answer:
We'll just do the casework. Let's call the racers Meg, Bob, X, Y, and Z.
If Meg finishes FIRST, we only have to order the other four cyclists. They can be ordered in 4!, or 24 ways.
If Meg finishes SECOND, she'll be ahead of Bob if Bob DOESN'T finish first. So we need
* X, Y, or Z to finish first
* then to order Bob and the other two racers (of X, Y, Z) who didn't win
We have 3 choices for the winner and 3*2*1 ways of arranging the others, for 3 * 3!, or 18 arrangements.
If Meg finishes THIRD, we need
* Two of X, Y, and Z to finish first and second
* then to order Bob and the last racer
We have 3*2 choices for the first two and 2*1 for the last two, for a total of 6*2, or 12 arrangements.
If Meg finishes FOURTH, we need Bob to finish last. So we only arrange the other three racers, for 6 total arrangements.
Obviously Meg can't finish last, so we have a total of 24 + 18 + 12 + 6 = 60 arrangements.
Last edited by
Matt@VeritasPrep on Sun Jul 06, 2014 6:28 pm, edited 1 time in total.
