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Question:

10 people, split into team of 6, team of 4=(10C4)=(10C6).

And into teams of 5,5= (10C5)/2

I don't understand why one divides by 2 in the second case. Say I choose am team of 6 by (10C6), whatever left is a team of 4. Why is the same not true for choosing a team 5?

Abhishek Bhatia
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    Sometimes looking at a smaller example makes it easier to see what is going on. $3$ people, split into team of $2$ and team of $1,$ the number of ways is $\binom32\binom11=3\cdot1=3.$ Split them into two teams of $1,$ the number of ways is $\binom31\binom21/2=3\cdot2/2=3.$ Why? – bof Apr 30 '16 at 23:28
  • @bof why did you divide by 2? – Abhishek Bhatia Apr 30 '16 at 23:36
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    Do you think $6$ is the right answer to the last problem in my common? If you have $3$ people, call them $A,B,C,$ there are $6$ ways to choose two teams of $1$? OK, let's list them.1. A&B 2. A&C 3. B&A. 4. B&C. 5. C&A. 6. C&B. The reason I divided by two is that I didn't think A&B and B&A should be counted as two different ways of making two teams of one. Of course, if they are to be counted as different ways, then I don't divide by $2$ and the answer if $6.$ – bof Apr 30 '16 at 23:45
  • @bof Thanks so much! Understood. – Abhishek Bhatia Apr 30 '16 at 23:53

1 Answers1

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This is the difference between:

  • split into two teams of 5; and:
  • split into team A (with 5 persons) and team B (also 5 persons)

In the first case, you don't know which team is which, so you have only one choice for a pair of choices from the second case (i.e. half the number of choices).
Which solution is correct depends on the context. It is not always right, to divide by two.

Ilja
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  • My case is the first one, I understand why division by 2 is there? – Abhishek Bhatia Apr 30 '16 at 23:36
  • you dont understand? Well, the second case is analogous to the 6+4-splitting, and in the first you have to count every pair of possibilities as one, because you dont know how to order the teams – Ilja Apr 30 '16 at 23:40