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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Dog Show - Permutation
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ayankm
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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:
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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vvijay146
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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?
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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DeepthiRajan
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liferocks
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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?
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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Stuart@KaplanGMAT
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You're missing that 6P3 doesn't equal 240!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:
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.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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Jeff@TargetTestPrep
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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.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
Answer: B
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