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How many possible orders

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How many possible orders

by phanikpk » Sun Jul 06, 2014 6:20 pm
Dear Experts, Please help

Meg and Bob among the five participants in a cycling race. If each participant finishes the race so that no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of bob?

Thanks in advance
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by Matt@VeritasPrep » Sun Jul 06, 2014 6:23 pm
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.
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by Brent@GMATPrepNow » Sun Jul 06, 2014 6:27 pm
phanikpk wrote:Dear Experts, Please help

Meg and Bob among the five participants in a cycling race. If each participant finishes the race so that no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of bob?

Thanks in advance

We can arrange the 5 people in 5! ways (= 120 ways).
Notice that, for HALF of these arrangements, Bob will be ahead of Meg.
For the OTHER HALF, Meg will be ahead of Bob.

So, the number of arrangements where Meg finishes ahead of Bob = 120/2 = 60
Here's are two similar questions:
https://www.beatthegmat.com/counting-six ... 47167.html
https://www.beatthegmat.com/permutation-t261691.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATinsight » Mon Jul 07, 2014 12:17 am
phanikpk wrote:
Dear Experts, Please help

Meg and Bob among the five participants in a cycling race. If each participant finishes the race so that no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of bob?

Thanks in advance
Hi Phanikpk,

Brent has mentioned the best method which is 5!/2 = 120/2 = 60

However another method that can just add another dimension to look at this problem is in the following three steps

Step 1: We can select two places out of 5 for selecting the places to be occupied by Meg and Bob, which can be selected in 5C2 ways = 10 ways

Step :2 In every selection of two places, Meg and Bob can be arranged only in 1 way as Meg should always finish before Bob as required in the question

Step :3 Remaining 3 can be arranged on three UNSELECTED places in 3! ways

Therefore total ways to make sure that Meg finishes befre Bob = 5C2 x 1 x 3! = 10 x 6 = 60 Answer
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