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In a race C, seedings are determined as follows: seeds 1, 2, and 3 are

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In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 03 Aug 2019, 01:16
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In a race C, seedings are determined as follows: seeds 1, 2, and 3 are given to the top three finishers in race A, while seeds 4, 5, and 6 are given to the top three finishers in race B. If race A and race B both consist of six entrants, how many possible arrangements of seeds are there for race C?

A) 12,200
B) 14,400
C) 120
D) 720
E) 16,000

Source: Math Bible (Jeff Sacksmann)

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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 03 Aug 2019, 02:12
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No. of possibile combinations for top 3 in race A is 6C3=20 and each of these 20 combinations can finish in 3! ways (arrangements of 1st 2nd and 3rd positions).

So the number of arrangements for first 3 places in race A is 6C3*3!=120 ways

Similarly we have 120 ways in race B

So the number of possibile arrangements for race C is 120*120=14400

Answer is (B)

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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 03 Aug 2019, 07:57
DarkHorse2019 wrote:
In a race C, seedings are determined as follows: seeds 1, 2, and 3 are given to the top three finishers in race A, while seeds 4, 5, and 6 are given to the top three finishers in race B. If race A and race B both consist of six entrants, how many possible arrangements of seeds are there for race C?

A) 12,200
B) 14,400
C) 120
D) 720
E) 16,000

Source: Math Bible (Jeff Sacksmann)


given: race A and B have 6 entrants each, and prizes are given to the top 3 of each;
then the num of arrangements of race C is the total ways of choosing top 3 and arranging their places.

race A: ways choosing top 3 from 6 = 6C3 = 20 AND ways arranging top 3 places = 3! = 6
race B: ways choosing top 3 from 6 = 6C3 = 20 AND ways arranging top 3 places = 3! = 6
race C: 20 * 6 * 20 * 6 = 400 * 36 = 14,400

Answer (B)
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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 03 Aug 2019, 08:24
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Seed 1- 6 possible from A
Seed 2- 5 possible from A
Seed 3- 4 possible from A
Seed 4- 6 possible from B
Seed 5- 5 possible from B
Seed 6- 4 possible from B

There are (6*5*4)*(6*5*4) = 120*120 = 14,400 possible arrangements.

Answer is (B)

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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 04 Aug 2019, 02:29
DarkHorse2019 wrote:
In a race C, seedings are determined as follows: seeds 1, 2, and 3 are given to the top three finishers in race A, while seeds 4, 5, and 6 are given to the top three finishers in race B. If race A and race B both consist of six entrants, how many possible arrangements of seeds are there for race C?

A) 12,200
B) 14,400
C) 120
D) 720
E) 16,000

Source: Math Bible (Jeff Sacksmann)


Official Solution:

Any of the six entrants in race A could be seeded 1st, leaving 5 possible 2nd seeds and 4 possible 3rd seeds. Starting over: any of the six entrants in race B could be seeded 4th, leaving 5 possible 5th seeds and 4 possible 6th seeds. Thus, the number of permutations is:

6 x 5 x 4 x 6 x 5 x 4 = 14,400

Option B is the correct answer
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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 04 Aug 2019, 20:34
firas92 wrote:
No. of possibile combinations for top 3 in race A is 6C3=20 and each of these 20 combinations can finish in 3! ways (arrangements of 1st 2nd and 3rd positions).

So the number of arrangements for first 3 places in race A is 6C3*3!=120 ways

Similarly we have 120 ways in race B

So the number of possibile arrangements for race C is 120*120=14400

Answer is (B)

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Hey, I directly applied Permutation and got the same answer. Please tell me why did u select combination first. I did 6P3*6P3 and got the same answer
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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are  [#permalink]

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New post 05 Aug 2019, 04:11
Shef08 wrote:
firas92 wrote:
No. of possibile combinations for top 3 in race A is 6C3=20 and each of these 20 combinations can finish in 3! ways (arrangements of 1st 2nd and 3rd positions).

So the number of arrangements for first 3 places in race A is 6C3*3!=120 ways

Similarly we have 120 ways in race B

So the number of possibile arrangements for race C is 120*120=14400

Answer is (B)

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Hey, I directly applied Permutation and got the same answer. Please tell me why did u select combination first. I did 6P3*6P3 and got the same answer


Hi Shef08,
its the same thing 6P3 = 6c3*3!
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Re: In a race C, seedings are determined as follows: seeds 1, 2, and 3 are   [#permalink] 05 Aug 2019, 04:11
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