It is currently 25 Nov 2017, 02:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Eight Alaskan Huskies are split into pairs to pull one of

Author Message
Director
Joined: 08 Jul 2004
Posts: 595

Kudos [?]: 283 [0], given: 0

Eight Alaskan Huskies are split into pairs to pull one of [#permalink]

### Show Tags

05 Oct 2004, 17:16
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Eight Alaskan Huskies are split into pairs to pull one of four sleds in a race. How many
different assignments of Huskies to sleds are possible?

OA to follow.
S

Kudos [?]: 283 [0], given: 0

Intern
Joined: 04 Oct 2004
Posts: 12

Kudos [?]: [0], given: 0

### Show Tags

05 Oct 2004, 17:42
If I understood correctly, the question asks for the number of 2 people combinations out of 8 total. So:
n=8, r=2 -> 8!/2!(8-2)!=28

Combinations/Permutations is one of my weak areas, so please let me know if this is correct.

-Irene

Kudos [?]: [0], given: 0

Director
Joined: 20 Jul 2004
Posts: 590

Kudos [?]: 162 [0], given: 0

### Show Tags

05 Oct 2004, 19:55
For m total numbers, dividing them into n distinct and equal groups of r each,
=m!/ (r!)^n
= 8!/(2!)^4
= 2520

Kudos [?]: 162 [0], given: 0

Intern
Joined: 04 Oct 2004
Posts: 12

Kudos [?]: [0], given: 0

### Show Tags

05 Oct 2004, 20:38
Hmm... I guess I conviniently ignored the 4 possible sleds part...(probably to make it simpler for myself)

Kudos [?]: [0], given: 0

Director
Joined: 08 Jul 2004
Posts: 595

Kudos [?]: 283 [0], given: 0

### Show Tags

06 Oct 2004, 03:52
hardworker, do the group has to be distinct ? how do you decide that?
S

Kudos [?]: 283 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 153 [0], given: 0

### Show Tags

06 Oct 2004, 04:02
Saurya,

I am soory for my wrong post about distinct or not groups... Just bad typing...

In this case groups are distincts cause there is 1 elected to race and 3 which will not take part of the race.

Another way to reach Hardworker result is the basic split calculation :
8C6.6C4.4C2 = 2520 which proves that hardworker also considered them as dinctinct.

Kudos [?]: 153 [0], given: 0

Director
Joined: 08 Jul 2004
Posts: 595

Kudos [?]: 283 [0], given: 0

### Show Tags

06 Oct 2004, 04:33
Twixt perhaps I am n ot getting your point. Look there are 4 groups. Now any one of the group will pull, So I guess there is no distinction between the groups. had it been 1,2,3 and 4, and if were asked one group to pull, then they would be distinct.
This is driving me insane!!
S

Kudos [?]: 283 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 153 [0], given: 0

### Show Tags

06 Oct 2004, 04:48
I am sorry I am afraid it was a second poor post anyway my last point il still available...

Distinct / not distinct : I did not properly read the question, I think there is no distinction between the groups here as we just know there are 4 sleds. I previously thought just 1 of them will race.

With no distinction the answer is 2520/4! = 105

Kudos [?]: 153 [0], given: 0

Director
Joined: 08 Jul 2004
Posts: 595

Kudos [?]: 283 [0], given: 0

### Show Tags

06 Oct 2004, 15:20
Well the OA is 2520 by Princeton. Now, can somebody help me to sort out this, more for conceptual reason.
S

Kudos [?]: 283 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 153 [0], given: 0

### Show Tags

06 Oct 2004, 15:26
To be honest I am really puzzled...

Despite PR OA I still think there is no distinction here, anyway Hardworker, Venksune, Paul or another guru is welcome to bring his/her opinion...

PS : before your posts I thought I was OK with this prob questions ! Grrrrrrrrrrrr

Kudos [?]: 153 [0], given: 0

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4284

Kudos [?]: 537 [0], given: 0

### Show Tags

06 Oct 2004, 16:36
ways to select a pair out of 8 people for sled#1: 8C2
ways to select a pair out of 6 people for sled#2: 6C2
ways to select a pair out of 4 people for sled#3: 4C2
ways to select a pair out of 2 people for sled#4: 2C2

8c2 * 6c2 * 4c2 * 2c2 = 2520

The sleds are distinct because it's one of 4 sleds in a race. Hence, you don't divide by 4! I don't think you would like your winning sled to be mixed up with another number as it wins the race
_________________

Best Regards,

Paul

Kudos [?]: 537 [0], given: 0

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4284

Kudos [?]: 537 [0], given: 0

### Show Tags

06 Oct 2004, 17:23
hardworker_indian wrote:
For m total numbers, dividing them into n distinct and equal groups of r each,
=m!/ (r!)^n
= 8!/(2!)^4
= 2520

Cool formula hardworker. I never used it and I'll keep this one in mind
_________________

Best Regards,

Paul

Kudos [?]: 537 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 153 [0], given: 0

### Show Tags

07 Oct 2004, 09:38
OUf...

I breath better, My first understanding of the problem was good...

Kudos [?]: 153 [0], given: 0

07 Oct 2004, 09:38
Display posts from previous: Sort by