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Eight dogs are in a pen when a sled owner comes to choose four dogs to
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28 Sep 2017, 01:55
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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
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28 Sep 2017, 05:06
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
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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
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28 Sep 2017, 07:05
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 Answer: Option D
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to
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03 Oct 2017, 09:28
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!(84)!] = (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: AB, Row 2: CD 2) Row 1: AC, Row 2: BD 3) Row 1: AD, Row 2: BC 4) Row 1: BC, Row 2: AD 5) Row 1: BD, Row 2: AC 6) Row 1: CD, Row 2: AB 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. Answer: D
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to
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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!(84)!] = (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: AB, Row 2: CD 2) Row 1: AC, Row 2: BD 3) Row 1: AD, Row 2: BC 4) Row 1: BC, Row 2: AD 5) Row 1: BD, Row 2: AC 6) Row 1: CD, Row 2: AB 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. Answer: D 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?



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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to
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07 Oct 2017, 06:13
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 itSTEP I; make teams of 4choosing 4 out of 8 is 8C4=\(\frac{8!}{4!4!}\) TIME saver dont simplify as more steps are involvedSTEP II : Make different pairs in 4choosing 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 waysso 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
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07 Aug 2019, 02:42
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!(84)!] = (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: AB, Row 2: CD 2) Row 1: AC, Row 2: BD 3) Row 1: AD, Row 2: BC 4) Row 1: BC, Row 2: AD 5) Row 1: BD, Row 2: AC 6) Row 1: CD, Row 2: AB 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. Answer: D 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
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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
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