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Dog Show - Permutation

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Dog Show - Permutation

by pkw209 » Tue Apr 20, 2010 4:09 pm

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At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first, second, and third place ribbons respectively, with no other dogs receiving a prize, how many different rosters of winners are possible for the two divisions together?

A) 28,800

B) 14,400

C) 720

D) 400

E) 36
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by ayankm » Tue Apr 20, 2010 7:32 pm
In the beagle group 3 dogs can receive 1st, 2nd or 3rd ribbons in 6P3 ways, i.e. 240.
Same for the dalmatian group.
For each combination in beagle group we can have 6P3 combination in dalmatian group.
So totally that becomes 57600 which incidentally is double for option A.

Am I missing something here? :roll:
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by vvijay146 » Tue Apr 20, 2010 8:09 pm
As we have to select 3 dogs from a larger set of 6 dogs, we have to use "Combinations" - n!/k!(n-k)! = 6!/3!3! = 20

Beagle = 20, Dalmation= 20,

Hence total roster winners possible are - 20 * 20 = 400, So the answer is D.

Sounds right?
Vijay
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by DeepthiRajan » Tue Apr 20, 2010 9:08 pm
Different rosters....Shouldn't that be Permutations instead
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by ansumania » Tue Apr 20, 2010 9:41 pm
what is OA
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by liferocks » Wed Apr 21, 2010 1:58 am
I am getting the ans as B 14400

number of roster of winner for beagle division 6C3*3!=120
number of roster of winner for dalmatian division 6C3*3!=120

so total number of different rosters of winners possible for the two divisions together are
120*120=14400

what is the OA?
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by pops » Wed Apr 21, 2010 2:28 am
out of 6 dogs:
1st: 6 ways
2nd: 5 ways
3rd: 4 ways

hence: 6*5*4

for both the groups combined=(6*5*4)^2=14,400
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by akhpad » Wed Apr 21, 2010 2:52 am
6P3 * 6P3 = 14400
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by eaakbari » Wed Apr 21, 2010 7:48 am
IMO B
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by Stuart@KaplanGMAT » Wed Apr 21, 2010 9:28 am
ayankm wrote:In the beagle group 3 dogs can receive 1st, 2nd or 3rd ribbons in 6P3 ways, i.e. 240.
Same for the dalmatian group.
For each combination in beagle group we can have 6P3 combination in dalmatian group.
So totally that becomes 57600 which incidentally is double for option A.

Am I missing something here? :roll:
You're missing that 6P3 doesn't equal 240!

nPk = n!/(n-k)!

6P3 = 6!/3! = 6*5*4 = 120

120*120 = 14400

However, as pops notes, we can reason it out instead of using a formula. Most of the permutations/combinations questions that you'll see on the GMAT can be attacked via logic/common sense rather than formulae.
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by pkw209 » Wed Apr 21, 2010 3:49 pm
This one should have been fairly easy. Not that Stuart needs any confirmation but the answer is B.
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by Jeff@TargetTestPrep » Wed Jun 20, 2018 4:02 pm
pkw209 wrote:At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first, second, and third place ribbons respectively, with no other dogs receiving a prize, how many different rosters of winners are possible for the two divisions together?

A) 28,800

B) 14,400

C) 720

D) 400

E) 36
Since the order of finishing in each division is important, we use permutations. The number of ways in which the awards can be provided is 6P3 x 6P3 = 6!/3! x 6!/3! = 6 x 5 x 4 x 6 x 5 x 4 = 14,400.

Answer: B

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