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At a certain pet show, exactly 6 dogs are entered in the bea
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16 Jun 2018, 10:18
16 Jun 2018, 10:18
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68% (01:26) correct
32% (01:27) wrong based on 50 sessions
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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 ways can ribbons be awarded to winners in the two divisions together?
A. 28,800
B. 14,400
C. 720
D. 400
E. 36
A. 28,800
B. 14,400
C. 720
D. 400
E. 36
Re: At a certain pet show, exactly 6 dogs are entered in the bea
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21 Jun 2018, 23:24
21 Jun 2018, 23:24
1
here is the answer
in the first group 3 dogs can receive the award in 6*5*4 = 120 ways
in the second group 120 ways
total ways = 120*120= 14400
in the first group 3 dogs can receive the award in 6*5*4 = 120 ways
in the second group 120 ways
total ways = 120*120= 14400
Re: At a certain pet show, exactly 6 dogs are entered in the bea
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27 Oct 2020, 01:41
27 Oct 2020, 01:41
1
In the first group, winner group containing 3 dogs can be formed in 6C3 ways - prizes can now be distributed in group 1 in 6C3*3! ways = 120.
In the second group, winner group containing 3 dogs can be formed in 6C3 ways - prizes can now be distributed in group 2 in 6C3*3! ways = 120.
Total no of ways in which prizes can be distributed in group 1 & group 2 together = 120 * 120 = 14400
In the second group, winner group containing 3 dogs can be formed in 6C3 ways - prizes can now be distributed in group 2 in 6C3*3! ways = 120.
Total no of ways in which prizes can be distributed in group 1 & group 2 together = 120 * 120 = 14400
