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Comnination/Permutaion Help Please

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Comnination/Permutaion Help Please

by bloomfield08 » Thu Apr 24, 2008 7:30 am
Eight dogs are in a pen when a sled owner comes to choose six dogs to form a sled team. If the dogs are to be paired in three rows of two dogs each and each variation in pairing is considered a different team, how many different sled teams can the owner form?
Top ten or bust...
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by netigen » Thu Apr 24, 2008 6:41 pm
first set --> 8P2
second set --> 6P2
third set --> 4P2

8P2 x 6P2 x 4P2 = ans = 21160

does that match to the OA?
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by mandy12 » Fri Apr 25, 2008 4:27 am
The answer is 8C6*6! = 20160

You can choose 6 dogs out of 8 in 8C6 ways ...and then arrange these 6 dogs in 6! ways
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by bloomfield08 » Tue Apr 29, 2008 5:20 am
What?
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by ajguerre » Thu May 01, 2008 5:45 pm
You cannot multiply by 6! since the pair AB is the same as BA (is A and B were dog 1 and 2 for example).

So in order to get the total number of pairings you first need to get the total number of combinations of the six dogs (i.e. 8C6) which equals to 28.

Then since you have 3 pairs per combination you multiply by 3!

Thus the answer is 28*3! = 168
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by ajguerre » Thu May 01, 2008 5:46 pm
You cannot multiply by 6! since the pair AB is the same as BA (if A and B were dog 1 and 2 for example).

So in order to get the total number of pairings you first need to get the total number of combinations of the six dogs (i.e. 8C6) which equals to 28.

Then since you have 3 pairs per combination you multiply by 3!

Thus the answer is 28*3! = 168
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by beeparoo » Mon May 26, 2008 12:46 pm
ajguerre wrote:You cannot multiply by 6! since the pair AB is the same as BA (if A and B were dog 1 and 2 for example)...
But the question states, "each variation in pairing is considered a different team", which tells me that team A-B is different than team B-A.

What is the OA??

Sandra
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by chidcguy » Mon May 26, 2008 1:32 pm
Here is what I did. Can the OP tell us what the correct answer is?

6 out of the 8 dogs can be chosen in 8 C 6 ways.

Out of 6 dogs chosen, 2 can be chosen in 6 C 2 ways and the 2 can be arranged in 2 ways in between them ie 6 C 2 X 2

out of the 4 dogs remaining 2 can be chose in 2 C 2 ways and the 2 can be arranged in 2 ways between them ie 4 C 2 X 2

similarly 2 C 2 X 2 ways for the final two

8 C 6 X ( 6 C 2 X 2 X 4 C 2 X 2 X 2 C 2 X 2)

In other words this is also equal to 8 C 6 X 6 P 2 X 4 P 2 X 2 P 2
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