Counting unordered items

lindah · 2010-01-25 06:27:08

Hi there,

I have an issue with the following question which I hope someone can help me with:

Nine players are to be divided into 2 teams of four and one umpire.
If two particular people cannot be on the same team together, how many different combinations are possible?

My approach was to Specify person A & B and derive the combinations

1) Person A is the umpire

2) Person B is the umpire

3) Person A is already chosen in the first team.
Person B cannot be chosen as one of the next three people, leaving

I add these combinations and get 175 - but this is incorrect.

Thanks in advance for any feedback!!

Last edited by lindah (2010-01-25 07:09:29)

Fruityloop · 2010-01-25 06:53:14

I believe your problem lies in step 3. You have to take into account that any 1 of the remaining 4 people can be the umpire so it should be...

adding them up gives 280 as the answer.
Is this correct?

lindah · 2010-01-25 07:16:12

Hi FL,

Thanks for your prompt response
The answer is 210
Your answer made me think Hi FL,

because I have accounted for person A & B in that position in steps 1) & 2)??

mathsyperson · 2010-01-25 08:00:59

I'd do this by finding the total amount of combinations, then subtracting the ones where A and B are in the same team.

Total : 9 x 8C4 = 630.
A and B in one team : 2 x 7 x 6C2 = 210.

Final answer : 420.

Fruityloop · 2010-01-25 08:01:07

Hmmm...
I'm not sure why that's the answer. In step 3, we aren't counting A or B as being the umpire. We are only counting the remaining 4 people, none of whom are A or B.
We can also count the remaining 7 people in turn as being the umpire.
We put A on one team and B on the other team so we have...

I get the same answer as before.

Last edited by Fruityloop (2010-01-25 08:02:49)

lindah · 2010-02-09 16:08:42

Thanks FL

Yes I keep coming up with the same conclusion as you 280
What I would do for some worked solutions

Math Is Fun Forum

#1 2010-01-25 06:27:08

Counting unordered items

#2 2010-01-25 06:53:14

Re: Counting unordered items

#3 2010-01-25 07:16:12

Re: Counting unordered items

#4 2010-01-25 08:00:59

Re: Counting unordered items

#5 2010-01-25 08:01:07

Re: Counting unordered items

#6 2010-02-09 16:08:42

Re: Counting unordered items

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