Counting arrangments

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bamboole: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sun Mar 16, 2008 8:09 pm

Counting arrangments

by bamboole » Wed Oct 01, 2008 12:58 am

Hi all,

I found a question in a Princton Review brochure:

Carmelo and LeBron participate in a seven-person footrace. If all seven contestants finish and there are no ties, how many different arrangements of finishes are there in which Carmelo defeats Le Bron?

A 5040
B 2520
C 720
D 120
E 42

Thanks for explaining your answer.

Regards,

a guest

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Source: Beat The GMAT — Problem Solving | Wed Oct 01, 2008 12:58 am

4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 am; Thanked: 27 times; Followed by:1 members; GMAT Score:660(

by 4meonly » Wed Oct 01, 2008 6:21 am

I think B - 2520
Correct?

Carmelo and LeBron participate in a seven-person footrace. If all seven contestants finish and there are no ties, how many different arrangements of finishes are there in which Carmelo defeats Le Bron?
My reasoning:

1st version
Carmelo (C) and LeBron (L)
C 1st others in 6! = 720

2nd version
C second, others 5*5!=600
E.G.:
if C is second, then 1st place can be filled by 5 people (L will not overtake C), for 3rd place - 5 people, 4th - 4 people, 5th - 3 people, 6th - 2 people, 7th-1
5*5*4*3*2*1 = 5*5! = 600

3rd version
if C was 3rd
for 1st place - 5
2nd- 5
4th-4, 5rd-3, 6th-2, 7th-1
5*4*4*3*2*1 = 4*5!

Same logic for C on 4th, 5th and 6th place

Finally, we get
6!+5*5! + 4*5! + 3*5!+2*5! + 5! = 2520

OA?

Last edited by 4meonly on Wed Oct 01, 2008 6:31 am, edited 1 time in total.

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Morgoth: Master | Next Rank: 500 Posts; Posts: 316; Joined: Mon Sep 22, 2008 12:04 am; Thanked: 36 times; Followed by:1 members

Re: Counting arrangments

by Morgoth » Wed Oct 01, 2008 6:29 am

bamboole wrote:Hi all,

I found a question in a Princton Review brochure:

Carmelo and LeBron participate in a seven-person footrace. If all seven contestants finish and there are no ties, how many different arrangements of finishes are there in which Carmelo defeats Le Bron?

A 5040
B 2520
C 720
D 120
E 42

Thanks for explaining your answer.

Regards,

a guest

IMO 2520

C - Carmelo and L - LeBron

CL12345

L finishes at 7th place - C12345 can be arranged in 6! ways = 720
L finishes at 6th place - C12345 can be arranged in 5*5! ways = 600
L finishes at 5th place - C12345 can be arranged in 4*5! ways = 480
L finishes at 4th place - C12345 can be arranged in 3*5! ways = 360
L finishes at 3th place - C12345 can be arranged in 2*5! ways = 240
L finishes at 2th place - C12345 can be arranged in 1*5! ways = 120

720+600+480+360+240+120 = 2520

This is long way but good for explanation.

Otherwise you could have done it more simply.

Alternate Method----------------------------------

CL12345 = all these can be arranged in 7! ways = 5040

exactly half of the times C will be ahead of L , 5040/2 = 2520

OA?

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mental: Master | Next Rank: 500 Posts; Posts: 145; Joined: Mon Sep 29, 2008 1:14 am; Thanked: 13 times

by mental » Wed Oct 01, 2008 11:06 pm

somewhere in between
7 slots, 2 people - totel arrangements = 7P2 = 42
out of that 50% of times C stays ahead of L
ie 42/2 = 21

rest 5 can be arranged in 5! ways = 120

21*120 = 2520