combinatorics help needed

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??

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.

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?

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?

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.

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.