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I am posting the good ones for your practice.. 4. Eight

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Senior Manager
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I am posting the good ones for your practice.. 4. Eight [#permalink]

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New post 03 Jul 2006, 19:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I am posting the good ones for your practice.. :)

4. 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?

(A) 32
(B) 64
(C) 420
(D) 1680
(E) 2520

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New post 03 Jul 2006, 21:25
Total combinations of different pairs = 8C2*6C2*4C2*2C2/4! = 105

Four sleds. Then total assigmments = 105 * 4 = 120
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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New post 04 Jul 2006, 19:30
ps_dahiya wrote:
Total combinations of different pairs = 8C2*6C2*4C2*2C2/4! = 105

Four sleds. Then total assigmments = 105 * 4 = 120



This is not even in answers.. I am attaching OE and OA.
Attachments

alaskan.JPG
alaskan.JPG [ 19.6 KiB | Viewed 650 times ]

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New post 04 Jul 2006, 19:35
sharadGmat wrote:
ps_dahiya wrote:
Total combinations of different pairs = 8C2*6C2*4C2*2C2/4! = 105

Four sleds. Then total assigmments = 105 * 4 = 120



This is not even in answers.. I am attaching OE and OA.


My bad on the calculations

This is 105 * 4! = 2520
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New post 04 Jul 2006, 19:43
# of ways to pick first pair of huskies = 8C2 = 28
# of ways to pick second pair of huskies = 6C2 = 15 ways
# of ways to pick third pair of huskies = 4C2 = 6
# of ways to pick last pair of huskies = 1

Total # of ways = 28*15*6 = 2520

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Re: alaskan huskies.. [#permalink]

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New post 05 Jul 2006, 06:43
sharadGmat wrote:
I am posting the good ones for your practice.. :)

4. 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?

(A) 32
(B) 64
(C) 420
(D) 1680
(E) 2520


good question but is it not necessary to specify that only 2 dogs pull 1 sled?

if all 8 pull 1 sledge and if pairs are of importance then there would be

7 x 5 x 3 x 4 = 420 possibilities.

Can the math Gods confirm this ?

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Re: alaskan huskies.. [#permalink]

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New post 05 Jul 2006, 12:04
old_dream_1976 wrote:
sharadGmat wrote:
I am posting the good ones for your practice.. :)

4. 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?

(A) 32
(B) 64
(C) 420
(D) 1680
(E) 2520


good question but is it not necessary to specify that only 2 dogs pull 1 sled?

if all 8 pull 1 sledge and if pairs are of importance then there would be

7 x 5 x 3 x 4 = 420 possibilities.

Can the math Gods confirm this ?



I think problem is more complicated.. The different teams possible =
(8C2*6C2*4C2*2c2)/4! .. Here we need to do 4! for the following reasons..

When we divide a,b,c,d into 2 teams there could be 4c2*2c2*=6 ways..
ie.
1. ab,cd
2. ac,bd
3 ad,bc
4. bc,ad same as 3
5. bd,ac same as 2
6. cd,bc same as 1

So we need to divide 6/2! to get the actual number of ways.. ie. 3.

Similarly for 4 teams , we need to divide b y 4!.

So in our case, there are total = (8C2*6C2*4C2*2c2)/4! =105 ways..

So these teams-ways or say team-sets be named as 1, 2, 3, 5 through 105 where each one has 4 teams..


Each team-set has 4 teams.. And these need to be assigned to 4 sleds..
So total ways = 105* 4(number of teams) P4(number of sledges) = 105*4!= 2520..

So in your case, if all the 8 dogs pull one sledge, then
105*4P1 = 105 * 4 = 420..


Making sense ?

Kudos [?]: 104 [0], given: 0

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Re: alaskan huskies.. [#permalink]

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New post 10 Jul 2006, 19:25
sharadGmat wrote:
old_dream_1976 wrote:
sharadGmat wrote:
I am posting the good ones for your practice.. :)

4. 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?

(A) 32
(B) 64
(C) 420
(D) 1680
(E) 2520


good question but is it not necessary to specify that only 2 dogs pull 1 sled?

if all 8 pull 1 sledge and if pairs are of importance then there would be

7 x 5 x 3 x 4 = 420 possibilities.

Can the math Gods confirm this ?



I think problem is more complicated.. The different teams possible =
(8C2*6C2*4C2*2c2)/4! .. Here we need to do 4! for the following reasons..

When we divide a,b,c,d into 2 teams there could be 4c2*2c2*=6 ways..
ie.
1. ab,cd
2. ac,bd
3 ad,bc
4. bc,ad same as 3
5. bd,ac same as 2
6. cd,bc same as 1

So we need to divide 6/2! to get the actual number of ways.. ie. 3.

Similarly for 4 teams , we need to divide b y 4!.

So in our case, there are total = (8C2*6C2*4C2*2c2)/4! =105 ways..

So these teams-ways or say team-sets be named as 1, 2, 3, 5 through 105 where each one has 4 teams..


Each team-set has 4 teams.. And these need to be assigned to 4 sleds..
So total ways = 105* 4(number of teams) P4(number of sledges) = 105*4!= 2520..

So in your case, if all the 8 dogs pull one sledge, then
105*4P1 = 105 * 4 = 420..


Making sense ?


thanks sharad I think i now understand the problem

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Re: alaskan huskies..   [#permalink] 10 Jul 2006, 19:25
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I am posting the good ones for your practice.. 4. Eight

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