GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
At a certain pet show, exactly 6 dogs are entered in the bea
[#permalink]
16 Jun 2018, 10:18
16 Jun 2018, 10:18
Expert Reply
1
Bookmarks
00:00
Question Stats:
67% (01:27) correct
32% (01:27) wrong based on 49 sessions
Hide Show timer Statistics
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
[#permalink]
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
[#permalink]
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
