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

lindah
Member
Registered: 2010-01-25
Posts: 121

Counting unordered items

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)

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

Fruityloop
Member
Registered: 2009-05-18
Posts: 143

Re: Counting unordered items

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?

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

lindah
Member
Registered: 2010-01-25
Posts: 121

Re: Counting unordered items

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)??

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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Counting unordered items

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.


Why did the vector cross the road?
It wanted to be normal.

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

Fruityloop
Member
Registered: 2009-05-18
Posts: 143

Re: Counting unordered items

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)

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

lindah
Member
Registered: 2010-01-25
Posts: 121

Re: Counting unordered items

Thanks FL

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

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