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Difficulty:
55%
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Question Stats:
65% (01:57) correct
35%
(01:49)
wrong
based on 181
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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. 120
B. 720
C. 12,200
D. 14,400
E. 16,000
A. 120
B. 720
C. 12,200
D. 14,400
E. 16,000
30 Dec 2021, 13:01
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Bunuel
Race C consists of two races race A and race B
Race A has 6 entrants so choosing top 3 from 6 can be done in 6C3 ways = 20
Now seeds can be distributed in 3*2*1 ways = 6 ways.
So total number of ways possible = 20*6 =120
Race B also has 6 entrants so here also selection can be done in 6C3ways = 20
Now seeds can be distributed in 3*2*1 ways = 6 ways.
So total number of ways possible = 20*6 =120
So race C has possible arrangements = 120*120 = 14400 D
06 Mar 2022, 11:30
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We can solve this Problem using Fundamental counting Way:
set 1 :
1 2 3
6 5 4
set 2:
4 5 6
6 5 4
it means for
1st Position we can choose in 6 ways from 6 participants
2nd Position we can choose in 5 ways from 5 participants(as 1st position is already taken i.e 6-1=5)
3rd position similar to 2nd position is chosen in 4 ways from 4 participants
so total for team 1 6X5X4 = 120
same has to be done for 2nd team which will yield 120
120*120 ==> 14400
set 1 :
1 2 3
6 5 4
set 2:
4 5 6
6 5 4
it means for
1st Position we can choose in 6 ways from 6 participants
2nd Position we can choose in 5 ways from 5 participants(as 1st position is already taken i.e 6-1=5)
3rd position similar to 2nd position is chosen in 4 ways from 4 participants
so total for team 1 6X5X4 = 120
same has to be done for 2nd team which will yield 120
120*120 ==> 14400
