Counting Principles - Help!

[Cari_R]

Can anyone help me with the following question?:

12 people need to be split up into teams for a quiz.

a) show that the number of ways of splitting them into two groups of the same size is 1/2 (12C6).


I first thought it was just 12C6 - you choose 6 for the first team and the rest go into the other one - but that is apparently not the right answer

Thank you for your help!

 

[ksdhart]

There's a trick here, involving the fact that the groups are the same size. Consider the following:

Say the first group you picked was (1, 2, 4, 5, 7, 9). Then the other group was (3, 6, 8, 10, 11, 12). But what if instead you had chosen (3, 6, 8, 10, 11, 12) as the first group? Then the second group would be (1, 2, 4, 5, 7, 9). What did you learn from this example? Can you apply the same logic to any other combinations of students in the first group?

 

[Cari_R]

Thank You!

Thanks! I got by now even though it took me longer than it should have :/ I did 12!/6!x6!x2! - I hope thats right...

 

[pka]

Can anyone help me with the following question?:
12 people need to be split up into teams for a quiz.
a) show that the number of ways of splitting them into two groups of the same size is 1/2 (12C6).
I first thought it was just 12C6 - you choose 6 for the first team and the rest go into the other one - but that is apparently not the right answer.

This in really not a combination problem, but it is an arrangement problem.

How many ways are there to arrange the letters in the word SOUTHERN? Ans is \(\displaystyle 8!\) because there are eight distinct letters.

How many ways are there to arrange the letters in the word MISSISSIPPI? Ans is \(\displaystyle \dfrac{11!}{(4!)^2(2!)}\) because there are eleven letters, BUT two letters are repeated four times and another is repeated twice.

Applying that logic how many ways can this string be rearrange AAAAAABBBBBB?
Ans: \(\displaystyle \dfrac{12!}{(6!)^2}\), there are twelve letters two of which are repeated six times.

Can you think of an A team and a B team? so that is the number of ways to form two NAMED teams.
But the question asked for two teams with no names.

So let's take care of that \(\displaystyle \dfrac{1}{2}\cdot\dfrac{12!}{(6!)^2}=\dfrac{1}{2} \cdot\dbinom{12}{6}\)

\(\displaystyle \dbinom{12}{6}\) is the combination of twelve taken six at a time.