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Meg and Bob are among the 5 participants in a cycling race. If each pa
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07 Jan 2008, 06:05
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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
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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26 Feb 2012, 23:01
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... I was able to recall several such questions from combinations: mothermarycomestome86407.html (or: maryandjoe126407.html); sixmobstershavearrivedatthetheaterforthepremiereofthe126151.htmlinhowmanydifferentwayscanthelettersaab91460.htmlgoldenrodandnohopeareinahorseracewith6contestants82214.htmlCheck our question banks for many more questions on combinations. PS: search.php?search_id=tag&tag_id=52DS: search.php?search_id=tag&tag_id=31Hope it helps
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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08 Jan 2008, 19:02
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! 24 + 18 + 12 + 6 = 60




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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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07 Jan 2008, 06:13
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
5!/2 = 60



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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08 Jan 2008, 18:18
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 ?



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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09 Jan 2008, 02:16
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 Answer = 24+18+12+6 = 60



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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09 Jan 2008, 21:11
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 24+18+12+6 = 60.



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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27 Sep 2009, 10:44
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



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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26 Feb 2012, 19:27
hmm, I went about it the long way until I read the response here.
When M finishes 1st: Permutations > (4P3) = 4! = 24
When M finishes 2nd: Permutations > (3P1 * 3P3) = 3 * 3! = 18 (3P1 ways of arranging people before M, 3P3 ways of arranging people after M)
When M finishes 3rd: Permutations > (3P2 * 2P2) = 6 * 2! = 12 (3P2 ways of arranging people before M, 2P2 ways of arranging people after M)
When M finishes 4th: Permutations > (3P3 * 1P1) = 3! * 1 = 6 (3P3 ways of arranging people before M, 1P1 ways of arranging people after M)
Total = 24 + 18 + 12 + 6 = 60.
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...



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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26 Feb 2012, 23:04
Also try Combinatorics chapter of Math Book to have an idea about the staff that is tested on the GMAT: mathcombinatorics87345.html Finally try Hard questions on combinations and probability with detailed solutions: hardestareaquestionsprobabilityandcombinations101361.htmlHope it helps.
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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27 Feb 2012, 05:07
fortsill wrote: hmm, I went about it the long way until I read the response here.
What happened with you, happens with many other people. I have given a detailed explanation of questions of this type in this post: http://www.veritasprep.com/blog/2011/10 ... spartii/
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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06 Jul 2013, 19:03
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
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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12 Mar 2015, 11:34
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
When I first saw this I thought combination but then realized its permutation because order does matter(Meg has to finish ahead of Bob). So we can use n!/(nk)! n represents the 5 participants and k would be the 3 remaining cyclers which the orders do not matter correct? so 5!/(53)!= 60 answer is C
Is this a correct approach?



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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12 Mar 2015, 21:46
GMAT01 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
When I first saw this I thought combination but then realized its permutation because order does matter(Meg has to finish ahead of Bob). So we can use n!/(nk)! n represents the 5 participants and k would be the 3 remaining cyclers which the orders do not matter correct? so 5!/(53)!= 60 answer is C
Is this a correct approach? I don't think so. You have used nPr = 5P3 which we use when we have to choose and arrange 3 objects out of 5. The logic here is a bit different. I suggest you to check out the solutions given above.
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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13 May 2015, 01:50
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 My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ?



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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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13 May 2015, 02:25
HarshitBShah wrote: 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 My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ? Hi HarshitBShah, You have considered one of the scenarios of the total possible combinations. When you assume Meg & Bob as one entity you only take the case where Bob is the next person behind Meg. Imagine a situation where Meg finishes 1st, since Bob has to finish behind Meg, Bob can finish at any place from 2nd to the 5th. Similarly when Meg finishes 2nd, Bob can finish at any place from 3rd to 5th and so on. Combining them as one entity would have worked had the question constrained Bob & Meg to finish the race in consecutive positions. Hope its clear Regards Harsh
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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13 May 2015, 22:06
HarshitBShah wrote: 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 My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ? "Meg finishes ahead of Bob" is different from "Bob finishes immediately after Meg". Meg could finish way ahead of Bob or just ahead. You accounted for cases such as: Meg, Bob, A, B, C and C, Meg, Bob, A, B etc but how about cases such as: Meg, B, Bob, A, C? You need to use the principle of symmetry here. Check out the link in gave in my post above.
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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05 Sep 2017, 18:06
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 We can create the following equation: total number of ways to finish the race = number of ways Meg finishes ahead of Bob + number of ways Meg does not finish ahead of Bob Since the total number of ways to complete the race is 5! = 120, and since there are an equal number of ways for Meg to finish ahead Bob as there are for her not to finish ahead of Bob, Meg can finish ahead of Bob in 60 ways. Alternate Solution: If Meg comes in the first, then Bob can finish the race in any of the remaining positions, so there are 4! = 24 ways this could happen. If Meg comes in the second, there are 3 choices for the first spot (since Bob can’t finish ahead of Meg) and the last 3 spots can be filled in 3! ways; so there are 3 x 3! = 3 x 6 = 18 ways this could happen. If Meg comes in the third, there are 3 choices for the first spot, 2 choices for the second spot, 2 choices for the fourth spot, and 1 choice for the last spot; so, this could happen in 3 x 2 x 2 = 12 ways. Finally, if Meg comes in the fourth, then Bob must finish last and the first 3 spots can be filled in 3! = 6 ways. Note that Meg cannot finish the race in the last position because then Bob will have finished the race ahead of Meg. In total, there are 24 + 18 + 12 + 6 = 60 ways Meg can finish the race ahead of Bob. Answer: C
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Re: Meg and Bob are among the 5 participants in a cycling race. If each pa
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21 Apr 2018, 07:16
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 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 Answer: C Cheers, Brent
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