4 baseball players each stand at different corners of

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

4 baseball players each stand at different corners of

by M7MBA » Sat Dec 23, 2017 3:41 am

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 OA is the option B.

I would like to know how to solve this PS question. Experts, may you give me an explanation? Thanks in advanced.

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Source: Beat The GMAT — Problem Solving | Sat Dec 23, 2017 3:41 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Dec 23, 2017 4:06 am

M7MBA 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 problem above is equivalent to the following:
How many ways can 4 players be arranged in a CIRCLE?

To count circular arrangements:
1. Place one person in the circle.
2. Count the number of ways to arrange the REMAINING people.

After one of the players has been placed in the circle, the number of ways to arrange the remaining 3 players = 3! = 6.

The correct answer is B.

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

by M7MBA » Sun Jan 07, 2018 11:33 am

Thank you so much GMATGuruNY. <i class="em em---1"></i>

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

4 baseball players each stand at different corners of

by regor60 » Mon Jan 08, 2018 6:51 am

M7MBA 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 OA is the option B.

I would like to know how to solve this PS question. Experts, may you give me an explanation? Thanks in advanced.

My approach is to first consider all the ways the 4 can be arranged, without limitation, which is 4! or 24.

Then consider that each arrangement has three look alikes that are different only because they are rotated 1 or more bases. Since their rpositions relative to each other don't change, they need to be removed.

So you remove three members in each group of 4. How many fours in 24 ? Six groups of 4. Remove 3 members from each, leaving one member in each of the 6 groups.

Of course, it is easier to just divide 24 by 4, the number of bases yielding 6

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Aug 12, 2019 10:43 am

M7MBA 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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