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combinatorics help needed

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by ashish1354 » Tue Sep 09, 2008 3:12 am
In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded the medals, then how many groups of medal winners are possible?

Since only 6 will advance to the final round i found no of groups possible for 3 out of 6 contestants i.e 6!/3!(6-3)!=20

but this answer is wrong can someone suggest what is wrong??
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Re: combinatorics help needed

by parallel_chase » Tue Sep 09, 2008 3:47 am
ashish1354 wrote:In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded the medals, then how many groups of medal winners are possible?

Since only 6 will advance to the final round i found no of groups possible for 3 out of 6 contestants i.e 6!/3!(6-3)!=20

but this answer is wrong can someone suggest what is wrong??
Out of 8, any 6 people can advance to the final round.
In final round any 3 out of 6 can be awarded medals.

Therefore,

Out of 8 any 3 can be awarded medals.

8C3 = 56.

Hope this helps.
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by bharathaitha » Tue Sep 09, 2008 5:24 am
six finalists can be selected from eight people in 8C6. Out of 6 finalists 3 people can be selected in 6C3 ways. Hence totla number of ways is

8C6*6C3.

Please let me know whether the above solution is correct or not.
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Re: combinatorics help needed

by Stockmoose16 » Tue Sep 09, 2008 2:57 pm
parallel_chase wrote:
ashish1354 wrote:In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded the medals, then how many groups of medal winners are possible?

Since only 6 will advance to the final round i found no of groups possible for 3 out of 6 contestants i.e 6!/3!(6-3)!=20

but this answer is wrong can someone suggest what is wrong??
Out of 8, any 6 people can advance to the final round.
In final round any 3 out of 6 can be awarded medals.

Therefore,

Out of 8 any 3 can be awarded medals.

8C3 = 56.

Hope this helps.
Wouldn't this need to be two different equations? First, you have to determine the number of finalists

8C6=28

Then, of those 28, how many ways can you arrange 3 winners:

28C3

Is this correct?
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Re: combinatorics help needed

by Stockmoose16 » Tue Sep 09, 2008 2:58 pm
parallel_chase wrote:
ashish1354 wrote:In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded the medals, then how many groups of medal winners are possible?

Since only 6 will advance to the final round i found no of groups possible for 3 out of 6 contestants i.e 6!/3!(6-3)!=20

but this answer is wrong can someone suggest what is wrong??
Out of 8, any 6 people can advance to the final round.
In final round any 3 out of 6 can be awarded medals.

Therefore,

Out of 8 any 3 can be awarded medals.

8C3 = 56.

Hope this helps.
Wouldn't this need to be two different equations? First, you have to determine the number of finalists

8C6=28

Then, of those 28, how many ways can you arrange 3 winners:

28C3

Is this correct?
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Re: combinatorics help needed

by parallel_chase » Tue Sep 09, 2008 3:57 pm
Wouldn't this need to be two different equations? First, you have to determine the number of finalists

8C6=28

Then, of those 28, how many ways can you arrange 3 winners:

28C3

Is this correct?
NO!

According to your working you are saying that there were 8 people competing for the semi finals and 28 people are competing in finals.

Does it sound right?
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Re: combinatorics help needed

by Stockmoose16 » Tue Sep 09, 2008 4:00 pm
parallel_chase wrote:
Wouldn't this need to be two different equations? First, you have to determine the number of finalists

8C6=28

Then, of those 28, how many ways can you arrange 3 winners:

28C3

Is this correct?
NO!

According to your working you are saying that there were 8 people competing for the semi finals and 28 people are competing in finals.

Does it sound right?
That's not what I'm saying at all. I'm saying that there are 8 people to start with, all vying for 6 spots. That means there are 8C6 possibilites=28 combinations. Of those 28 combinations, only 3 get a medal. Thus, 28C3. Sounds right to me.
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by Shrinidhi » Tue Sep 09, 2008 8:20 pm
bharathaitha wrote:six finalists can be selected from eight people in 8C6. Out of 6 finalists 3 people can be selected in 6C3 ways. Hence totla number of ways is

8C6*6C3.

Please let me know whether the above solution is correct or not.
I also think this is correct..
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Re: combinatorics help needed

by Shrinidhi » Tue Sep 09, 2008 8:22 pm
Stockmoose16 wrote:
parallel_chase wrote:
Wouldn't this need to be two different equations? First, you have to determine the number of finalists

8C6=28

Then, of those 28, how many ways can you arrange 3 winners:

28C3

Is this correct?
NO!

According to your working you are saying that there were 8 people competing for the semi finals and 28 people are competing in finals.

Does it sound right?
That's not what I'm saying at all. I'm saying that there are 8 people to start with, all vying for 6 spots. That means there are 8C6 possibilites=28 combinations. Of those 28 combinations, only 3 get a medal. Thus, 28C3. Sounds right to me.
Your statement "Of those 28 combinations, only 3 get a medal. Thus, 28C3. Sounds right to me." doesnt sound logical. Combinations dont get medals, people do.
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