Re: [Math4u] A combinations/probability problem

So, ordering is important in your problem?

 

any way, I guess it can be solved with ordering or without ordering. The answer is the same.

 

There is a formula for arrangements with repeatition Kirk.

As for arranging 1,1,2,2,3,3:

 

6! / (2! * 2! * 2!), that is 6!/8 which is 6*5*3 = 90 ways.

Kirk Bevins <hatton02@yahoo.co.uk> wrote:

OK. Imagine you are selecting 6 numbers from a random selection of 20
numbers. The numbers are:

1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10.

If anybody has seen the British TV show, Countdown, then this is
where the problem comes from.

How much more likely is it of choosing 1, 3, 4, 5, 6, 7 compared with
1, 1, 2, 2, 3, 3?

The next question (which probably helps solve the above) is:
There are 6! ways of arranging 1, 3, 4, 5, 6, 7 in some order. How
many ways are there of arranging 1, 1, 2, 2, 3, 3 in some order? This
kind of thing I struggle with as we have repeated numbers. Do we just
do 6! divided by 3 as there are 3 repeats? I don't have the right
intuition for combinations.

Any explanation as to how you worked these out would be appreciated.
I don't understand combinations so don't feel like you're patronising
if you explain fully!!

Kirk

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