Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Meg and Bob are among the 5 participants in a cycling race. [#permalink]
07 Jan 2008, 05:05

1

This post received KUDOS

9

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

62% (02:07) correct
38% (01:17) wrong based on 360 sessions

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

Re: still on perms and combs [#permalink]
07 Jan 2008, 05:13

2

This post received KUDOS

2

This post was BOOKMARKED

There are 5! possible ways for the 5 participants to finish the race without any ties. Logically, Meg will finish ahead of Bob in exactly half of these outcomes

Re: still on perms and combs [#permalink]
08 Jan 2008, 18:02

5

This post received KUDOS

1

This post was BOOKMARKED

pmenon wrote:

ok, this is how i started to think about it ... didnt get the right answer though

there are 5 spots, so meg could come in first, second, third, fourth or fifth

if she comes in first, then there are 4 spots behind her that bob could be in ... so thats 4! = 24 ways

if she comes in second, then there are 3 spots behind her ... bob could be in any of those, so thats 3! = 6 ways

if she was in third, only two spots behind her, so thats 2 ways

if she comes in fourth, only one way ... adding them up, i get 24+6+2+1 = 33 ways.

Where did i go wrong ?

look at the problem this way:

M _ _ _ _ = 4! ways as you said _ M _ _ _ = 3 ways for Bob, still 3 options for the next (because unlike Bob she can occupy the first slot), 2 options for the the fourth one and 1 for the last. So 3*3*2*1 which is 3*3! _ _ M _ _ = 2*3! _ _ _ M _ = 1*3!

Re: still on perms and combs [#permalink]
09 Jan 2008, 01:16

2

This post received KUDOS

1

This post was BOOKMARKED

marcodonzelli wrote:

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

A. 24 B. 30 C. 60 D. 90 E. 120

C.

M finishes first then B can finish 4 ways and other three can finish in 3! ways = 1*4*3! = 24 M finishes second then B can finish in 3 ways and other three can finish in 3! ways = 1*3*3! = 18 M finishes third then B can finish in 2 ways and other three can finish in 3! ways = 1*2*3! = 12 M finishes fourth then B can finish in 1 way and other three can finish in 3! ways = 1*1*3! = 6

Re: still on perms and combs [#permalink]
09 Jan 2008, 20:11

marcodonzelli wrote:

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

A. 24 B. 30 C. 60 D. 90 E. 120

MBXXX we have 4! possibilities = 24 XMBXX we have 3! ways to arrange the X's and 3 ways to arrange B so 3!*3 =18 XXMBX we have 3! ways to arrange the X's and 2 ways to arrange B. so 3!*2 =12 XXXMB only 3! =6

Re: still on perms and combs [#permalink]
27 Sep 2009, 09:44

1

This post received KUDOS

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

A. 24 B. 30 C. 60 D. 90 E. 120

Ans: C. 60 M stands for Meg If M takes first place, the other 4 can be arranged in 4! ways. If M takes second place, the other 4 can be arranged in 3 * 3! ways such that Bob finishes after Meg. If M takes third place, the other 4 can be arranged in 6 * 2! ways such that Bob finishes after Meg. If M takes fourth place, the other 4 can be arranged in 3! ways such that Bob finishes after Meg. M cannot take fifth place, because in tat case Bob will finish ahead of Meg.

So total number of possibilities is = 4! + 3 * 3! + 6 * 2! + 3! = 60

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
26 Feb 2012, 22:01

3

This post received KUDOS

Expert's post

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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? A. 24 B. 30 C. 60 D. 90 E. 120

Total # of ways the race can be finished is 5!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 5!/2=60.

Answer: C.

fortsill wrote:

That being said, the other technique of arriving at the same solution by 5!/2 is far more elegant and time saving!

Do you know of any other such examples where the problem seems a lot simpler than it really is...

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
06 Jul 2013, 18:03

Subject: Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

I solved in a different way: All the possible positions: 5! = 120 Restriction: we have to eliminate all the combinations in which Bob finishes ahead of Meg:

If Bob finishes 1st, (_ _ _ _ B) Meg could finish 2nd, 3th, 4th or 5th --> 4 possibilities. If Bob finishes 2nd, Meg finishes 3th, 4th or 5th. --> 3 possibilities The same for Bob 3th and 4th. Bob can not finishes 5th.

This sum up to 10 combinatios: 4 + 3 + 2 + 1 = 10.

In addition, we have the rest of the combinations of the of the positions remaining 3 competitors: 3!

Now, we have to combine the positions of Meg and Bob with the positions of the rest of the competitors:

3! * 10 = 60

Finally: 120 - 60 = 60 _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
30 Dec 2013, 05:09

marcodonzelli wrote:

Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and 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?

A. 24 B. 30 C. 60 D. 90 E. 120

By symmetry 5! /2

Meg will finish ahead of Bob in half of all the possible ways