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4 baseball players each stand at different corners of a baseball diamo
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08 Nov 2009, 13:02
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4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond? a. 4 B. 6 C. 16 D. 24 E. 256 I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.
OA = B
OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.
I get the factorial part, 4*3*2*1 = 24  but can't figure out why we are dividing by 4?
Thanks for any input.
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Re: 4 baseball players each stand at different corners of a baseball diamo
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08 Nov 2009, 14:50
ajthomas wrote: Before I start, I hope this in the right place. I got this question from another website.
Q: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
a) 4 b) 6 c) 16 d) 24 e) 256
I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.
OA = B
OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.
I get the factorial part, 4*3*2*1 = 24  but can't figure out why we are dividing by 4?
Thanks for any input. I don't know why there is such a solution in OE. For me it's circular permutation for 4 which is (41)!=3!=6. The number of circular permutations of n different objects is (n1)!. Each circular permutation corresponds to n linear permutations depending on where we start. Since there are exactly n! linear permutations, there are exactly n!/n permutations. Hence, the number of circular permutations is the same as (n1)!. When things are arranged in places along a line with first and last place, they form a linear permutation. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first or last place, they form a circular permutation. The permutation in a row or along a line has a beginning and an end, but there is nothing like beginning or end or first and last in a circular permutation. In circular permutations, we consider one of the objects as fixed and the remaining objects are arranged as in linear permutation. Thus, the number of permutations of 4 objects in a row = 4!, where as the number of circular permutations of 4 objects is (41)! = 3!.
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Re: 4 baseball players each stand at different corners of a base
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08 Aug 2013, 11:19
gmatpapa wrote: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
A. 4 B. 6 C. 16 D. 24 E. 256 it is like a circular arrangement. total ways for n people to arrange in a circle is = factorial (n1) in this case n= 4 hence ans = 3 factorial = 6 hence B
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Re: 4 baseball players each stand at different corners of a base
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24 Dec 2014, 14:57
Hi All, These types of permutation questions can sometimes be confusing, so you might find that a "visual component" will help you to stay organized and get to the correct answer. In this question, we have to place 4 players at 4 corners of a diamond (although you could just as easily have a question that places them at equidistant spots on a circle). If we were placing those 4 players in a straight line, then the number of possibilities would be a simple calculation: (4)(3)(2)(1) = 24 HOWEVER, since we're told that we can count an option ONLY if the relative positions of the players differ. This means that the following 4 options are all the SAME thing: ABCD BCDA CDAB DABC The first option (ABCD) is simply "revolved" around the diamond, so the relative positions of the players are NOT different. Mathematically, we have to divide 24 by 4, so that remove all of the "duplicates" Final Answer: All in all, this a relatively rare issue on the GMAT (you probably won't see it; even if you aced the Quant section, you still probably wouldn't see it). GMAT assassins aren't born, they're made, Rich
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Re: 4 baseball players each stand at different corners of a baseball diamo
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07 Jul 2015, 08:40
ajthomas wrote: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond? a. 4 B. 6 C. 16 D. 24 E. 256 I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.
OA = B
OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.
I get the factorial part, 4*3*2*1 = 24  but can't figure out why we are dividing by 4?
Thanks for any input. How do i recognize if it's about circular permutation? ) Are there any other good examples/theory about this topic?



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Re: 4 baseball players each stand at different corners of a baseball diamo
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07 Jul 2015, 08:45
reto wrote: ajthomas wrote: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond? a. 4 B. 6 C. 16 D. 24 E. 256 I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.
OA = B
OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.
I get the factorial part, 4*3*2*1 = 24  but can't figure out why we are dividing by 4?
Thanks for any input. How do i recognize if it's about circular permutation? ) Are there any other good examples/theory about this topic? Players make closed circle, so circular arrangement. Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.
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Re: 4 baseball players each stand at different corners of a baseball diamo
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24 Dec 2015, 12:31
Bunuel wrote: ajthomas wrote: Before I start, I hope this in the right place. I got this question from another website.
Q: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
a) 4 b) 6 c) 16 d) 24 e) 256
I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.
OA = B
OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.
I get the factorial part, 4*3*2*1 = 24  but can't figure out why we are dividing by 4?
Thanks for any input. I don't know why there is such a solution in OE. For me it's circular permutation for 4 which is (41)!=3!=6. The number of circular permutations of n different objects is (n1)!. Each circular permutation corresponds to n linear permutations depending on where we start. Since there are exactly n! linear permutations, there are exactly n!/n permutations. Hence, the number of circular permutations is the same as (n1)!. When things are arranged in places along a line with first and last place, they form a linear permutation. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first or last place, they form a circular permutation. The permutation in a row or along a line has a beginning and an end, but there is nothing like beginning or end or first and last in a circular permutation. In circular permutations, we consider one of the objects as fixed and the remaining objects are arranged as in linear permutation. Thus, the number of permutations of 4 objects in a row = 4!, where as the number of circular permutations of 4 objects is (41)! = 3!. HI BB, I'm not sure if the correct answer is 3!. Here is my understanding. Feel free to correct me. However, I am still unable to arrive at correct answer! In general, YES.. it is (n1)! incase of circular arrangement. But, Can a diamond (with equal sides) be circular? I guess, it can be considered a square arrangement. In such cases, should not it be 2 x (n1)! ie; 2 x (3!) = 12



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4 baseball players each stand at different corners of a baseball diamo
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28 Dec 2015, 19:28
madhusudhan237 wrote: HI BB,
I'm not sure if the correct answer is 3!. Here is my understanding. Feel free to correct me. However, I am still unable to arrive at correct answer!
In general, YES.. it is (n1)! incase of circular arrangement. But, Can a diamond (with equal sides) be circular? I guess, it can be considered a square arrangement. In such cases, should not it be 2 x (n1)! ie; 2 x (3!) = 12
Responding to a pm: It is not an international quality question and hence unsuitable for GMAT. Note that an international applicant may not know that a "baseball diamond" is supposed to be a square (the question doesn't say this). What one may assume is that it is a diamond (a kite or a rhombus). So there will be two types of corners in general and the first person can occupy the first place in 2 ways. The rest will occupy places in 3! ways and hence you might get the answer as 12. If it were specified that it is a square, then all corners would be the same and you would get the answer as 3!.
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Re: 4 baseball players each stand at different corners of a base
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12 Aug 2019, 10:57
gmatpapa wrote: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
A. 4 B. 6 C. 16 D. 24 E. 256 The sentence “Two arrangements of baseball players are considered different only when the relative positions of the players differ.” means that the arrangement ABCD is the same as BCDA, which in turn is the same as CDAB etc. So, let’s keep the player A fixed at the home plate and find the number of ways we can arrange players B, C and D on the three bases. We see that we can do this in 3! = 6 ways. Thus, there are 6 ways we can arrange the players around the diamond. Alternate Solution: This is a circular permutations problem, since the baseball diamond can be likened to a circular table with 4 players being seated. We thus use the formula (n  1)!, where n is the number of players. Thus, we have: (n  1)! = (4  1)! = 3! = 6 Answer: B
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Re: 4 baseball players each stand at different corners of a base
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