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# Eight dogs are in a pen when a sled owner comes to choose four dogs to

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Math Expert
Joined: 02 Sep 2009
Posts: 57191
Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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28 Sep 2017, 01:55
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Question Stats:

40% (01:57) correct 60% (01:59) wrong based on 89 sessions

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Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

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Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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28 Sep 2017, 05:06
1
The total number of ways of choosing 4 dogs from 8 is $$8C4 = \frac{8*7*6*5}{4*3*2} = 70$$

Since there are 4 dogs in order to place in 4 spots, and each pairing of 2 dogs is considered
a different team, there are a total of $$\frac{4!}{2!*2!} = 3*2 = 6$$ ways

Hence, the different sled teams the owner can have are $$70*6 = 420$$(Option D)
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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28 Sep 2017, 05:15
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

8c2*6c2\2
210(C)

not sure whether it is right or not.
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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28 Sep 2017, 07:05
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1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

First we need 4 dogs in total out of eight = 8C4

Now we need to just pick two dogs for the first row so that remaining two dogs will be automatically be selected for other row = 4C2

Total ways = 8C4 * 4C2 = 70*6 = 420

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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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03 Oct 2017, 09:28
1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.

1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B

Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.

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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.

1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B

Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.

Why have we done 4,5 and 6 selection?What i can discern from the question is different pairings form different teams. So if AB is paired once whether in row 1 or 2, its pairing is done.

Please explain what is wrong in my interpretation of the question?
Math Expert
Joined: 02 Aug 2009
Posts: 7755
Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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07 Oct 2017, 06:13
1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

Hi..

let's do the Q step by step and understand what is the TRICK in it

STEP I;-
make teams of 4
choosing 4 out of 8 is 8C4=$$\frac{8!}{4!4!}$$
TIME saver- dont simplify as more steps are involved

STEP II :-
Make different pairs in 4
choosing 2 out of 4 = 4C2 = $$\frac{4!}{2!2!}$$
The TRICKY part - when we choose two out of 4, the other 2 are already there as SECOND pair. BUT 4C2 counts them as 2 separate ways
so divide by 2!
so $$\frac{4!}{2!2!*2}$$
Also it is also valid for the second set of 4..

answer = $$\frac{8!}{4!4!}*\frac{4!}{2!2!*2}*2=\frac{8*7*6*5}{4}=7*6*5*2=420$$

D
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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07 Aug 2019, 02:42
1
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24
B. 70
C. 210
D. 420
E. 1,680

There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.

1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B

Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.

Hi,

I have a doubt, can i choose 8C2 (for 2 dogs) and then 6C2 (for remaining 2 dogs)? Hence =8c2* 6c2 (is this right ?)
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Eight dogs are in a pen when a sled owner comes to choose four dogs to  [#permalink]

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11 Aug 2019, 13:58
The way I have done was thinking: we have 8 dogs and we are going to select 4.

So the total number of choices would be: 8 * 7 * 6 * 5 (it reduces 1 each time since you chose n dogs already and can't select it again).
Then this total number of combinations has 2 double countings, since there are 2 groups that are analyzed as one "entity", then we have:

(8*7*6*5)/ (2*2) = 420.
Eight dogs are in a pen when a sled owner comes to choose four dogs to   [#permalink] 11 Aug 2019, 13:58
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