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there are 28 students .4 of them are choosen to be the first speaker ,second speaker ,third speaker and last speaker in the debate team

how many ways are there to form the debate team

 if John and Sam are already choosen to be tthe team debate , find the probability that they speck next to each other

Guest Mar 9, 2015

Best Answer 

 #1
avatar+87333 
+5

We have 28 ways to choose the first speaker, 27 ways to choose the second speaker, etc.

So...the total number of ways to choose 4 speakers is 28 x 27 x 26 x 25 = 491,400

 

For the second, we have 4! ways of arranging any 4 speakers = 24

Considering that John and Sam can be arranged two ways, either John first and Sam second, or vice-versa, and they can fill any of three positions.... (1,2) (2,3)(3,4)...then 2 x 3 = 6 arramgements. And for each of these the other two people can be arranged in two ways. So, 6 x 2  = 12 arrangements

So the probability that they speak next to each other is 12 arrangements/24 possible arrangements = 1/2 = 50%.

 

  

CPhill  Mar 9, 2015
 #1
avatar+87333 
+5
Best Answer

We have 28 ways to choose the first speaker, 27 ways to choose the second speaker, etc.

So...the total number of ways to choose 4 speakers is 28 x 27 x 26 x 25 = 491,400

 

For the second, we have 4! ways of arranging any 4 speakers = 24

Considering that John and Sam can be arranged two ways, either John first and Sam second, or vice-versa, and they can fill any of three positions.... (1,2) (2,3)(3,4)...then 2 x 3 = 6 arramgements. And for each of these the other two people can be arranged in two ways. So, 6 x 2  = 12 arrangements

So the probability that they speak next to each other is 12 arrangements/24 possible arrangements = 1/2 = 50%.

 

  

CPhill  Mar 9, 2015

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