Ways of making a team

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Ways of making a team

Postby prime » Sat Oct 27, 2007 3:47 am UTC

This was one of the questions at our last little local math meet thing:

If you have 5 boys and 5 girls, how many 4 person teams with at least 1 boy can you make?

I did it by (5*9*8*7)/4! but that is not the right answer. I do know the right answer and a method for getting it, but what I want to know is why my method doesn't work. Any ideas?

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Re: Ways of making a team

Postby ikerous » Sat Oct 27, 2007 4:02 am UTC

Edit: I had a buncha stuff here but it was crap. I wrote a script and got 4440
Last edited by ikerous on Sat Oct 27, 2007 5:05 am UTC, edited 4 times in total.

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Re: Ways of making a team

Postby aguacate » Sat Oct 27, 2007 4:45 am UTC

One way to do it is:

5C4 + (5C3*5C1) + (5C2*5C2) + (5C1*5C3)

where aCb is "a choose b" or a!/[(a-b)!*b!]. So it's:

5!/4! + 2*5!*5!/(4!*2!*3!) + [5!/(3!*2!)]^2

5 + 14400/144 + (120/12)^2

5 + 100 + 100

= 205

What is your method trying to do? Where did your numbers come from?

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Re: Ways of making a team

Postby mikegoo » Sat Oct 27, 2007 7:33 am UTC

This is a very common mistake many of my students make (and I made when I was a student).

A is a boy and 123 are girls. If we try to do a calculation where order matters and then divide by the number of repeats (4! for this case), we need to make sure our method gets all of the arrangments. A123 will be found with your calculation as will A132, A213, A231, A312, A321. None of the other arrangements of this particular group of 4 get counted by your method as a girl will never be placed in the first slot. This means dividing by 4! for the repeats will lower your count below the actual count.

The "nicest" method for this is actually along the lines of aquacate but with slightly fewer calculations.

Find the total number of teams ignoring gender then subtract the number of groups with all girls.
10C4 - 5C4 = 205

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