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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]  03 Jul 2006, 18:23
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
CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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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

Senior Manager
Joined: 20 Feb 2006
Posts: 331
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Kudos [?]: 18 [0], given: 0

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 [ 19.6 KiB | Viewed 468 times ]

CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 121 [0], given: 0

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.

This is 105 * 4! = 2520
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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# 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
Senior Manager
Joined: 09 Aug 2005
Posts: 286
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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 ?
Senior Manager
Joined: 20 Feb 2006
Posts: 331
Followers: 1

Kudos [?]: 18 [0], given: 0

old_dream_1976 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
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 ?
Senior Manager
Joined: 09 Aug 2005
Posts: 286
Followers: 1

Kudos [?]: 2 [0], given: 0

old_dream_1976 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
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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