Carmelo and LeBron participate in a seven-person footrace on

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Carmelo and LeBron participate in a seven-person footrace on the basketball court during All-Star Weekend. If all seven contestants finish (including Charles Barkley) and there are no ties, how many different arrangements of finishes are there in which Carmelo defeats LeBron?

(A) 5040
(B) 2520
(C) 720
(D) 120
(E) 42

OA B

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by [email protected] » Thu Oct 25, 2018 6:47 pm
Hi All,

We're told that Carmelo and LeBron participate in a seven-person footrace on the basketball court during All-Star Weekend, all seven contestants finish and there are no ties. We're asked for the number of different arrangements in which Carmelo defeats LeBron. This question is a Permutation with a small 'twist' to the math.

With 7 racers and no 'ties', there would be 7! = (7)(6)(5)(4)(3)(2)(1) = 5040 possible arrangements.

However, we're asked for the total number in which Carmelo finishes BEFORE LeBron. With any 2 racers, exactly HALF of the options would involve one racer finishing ahead of the other. Here, that would be 5040/2 = 2520

Final Answer: B

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by swerve » Fri Oct 26, 2018 9:25 am
Arranging 7 persons on 7 positions where specific one person is always ahead of the other specific = 7!/2 = 2520.

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by Scott@TargetTestPrep » Tue Oct 30, 2018 10:26 am
BTGmoderatorDC wrote:Carmelo and LeBron participate in a seven-person footrace on the basketball court during All-Star Weekend. If all seven contestants finish (including Charles Barkley) and there are no ties, how many different arrangements of finishes are there in which Carmelo defeats LeBron?

(A) 5040
(B) 2520
(C) 720
(D) 120
(E) 42
Since there are 7 participants, the total number of possible finishes is 7! = 5,040. Since the number of ways in which Carmelo defeats Lebron is the same number of ways in which Lebron defeats Carmelo, there are 5,040/2 = 2,520 ways in which Carmelo defeats Lebron.

Answer: B

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by Brent@GMATPrepNow » Tue Oct 30, 2018 12:15 pm
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