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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Permutations and combinations ##### This topic has 1 expert reply and 5 member replies ## Permutations and combinations Hello, Need help for the following questions. (not sure about the accuracy of answer choices). Just want to validate my method of solving this problem. Q. There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible? (A) 20 (B) 25 (C) 40 (D) 60 (E) 125 Regards MSD _________________ When the going gets tough, the tough gets going. Legendary Member Joined 20 Jun 2007 Posted: 1153 messages Followed by: 2 members Upvotes: 146 Target GMAT Score: V50 msd_2008 wrote: Hello, Need help for the following questions. (not sure about the accuracy of answer choices). Just want to validate my method of solving this problem. Q. There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible? (A) 20 (B) 25 (C) 40 (D) 60 (E) 125 Regards MSD 5 cars - 3 red, 1 yellow & 1 blue this can be set-up in an anagram form - RRRYB No. of ways 5 cars can be arranged in 5 parking spaces - 5! No. of ways cars could be repeated - 3! Total permutations = 5!/3! = 120/6 = 20 OA? _________________ No rest for the Wicked.... Senior | Next Rank: 100 Posts Joined 29 Jun 2008 Posted: 66 messages Upvotes: 1 Yes. The OA is indeed 20 i.e A However, I am not able to understand why did we divide 5! by 3!. Can you please explain a bit? Also, why cant we follow the following method 5c3 + 5c1 + 5c1 = 10 + 5 +5 = 20 Regards MSD _________________ When the going gets tough, the tough gets going. Legendary Member Joined 20 Jun 2007 Posted: 1153 messages Followed by: 2 members Upvotes: 146 Target GMAT Score: V50 msd_2008 wrote: However, I am not able to understand why did we divide 5! by 3!. Can you please explain a bit? Sure This question can be easily converted into anagram question How many ways can you arrange the word BAAAY. 5!/3! Now the reason why 3! is divided is because they we have 3 identical A's now if we dont divide it by 3!, it will take BA1A2A3Y different from BA2A1A3Y, which in actual terms we cannot, therefore we divide the total arrangements by the number of repetitions to avoid this confusion. Hope this helps. msd_2008 wrote: 5c3 + 5c1 + 5c1 = 10 + 5 +5 = 20 Regards MSD You cannot follow this, well to be honest I dont understand what you did. But all I can say is that this question is of permutations not combination. _________________ No rest for the Wicked.... Senior | Next Rank: 100 Posts Joined 29 Jun 2008 Posted: 66 messages Upvotes: 1 Parallel_chase, I understood your explaination and as a matter of fact I have solved questions similar to this....such as how many ways can the letters of the word APPLE be arranged. Here we say 5!/2! because of 2 P's. Dont know what I was thinking while approaching the question. Anyways, thanks for your help! Regards MSD _________________ When the going gets tough, the tough gets going. Newbie | Next Rank: 10 Posts Joined 26 Nov 2011 Posted: 4 messages I know the OA is A but I have a problem with this answer. The question specify that the cars must all be parked in the same direction. From my understanding that means that there are least two directions. So the final answer you get by using commutation should be multiplied by two. Wouldn't that make sense ? ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2444 messages Followed by: 18 members Upvotes: 43 msd_2008 wrote: Q. There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible? (A) 20 (B) 25 (C) 40 (D) 60 (E) 125 The cars can be displayed in the following number of ways, using the indistinguishable permutations formula: (5!)/(3!) = 5 x 4 = 20 ways. Note that we divided by 3! because the 3 red cars are identical to each other (i.e., they are indistinguishable). Answer: A _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. 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