[Math4u] Re: A combinations/probability problem

There are only a few rules used for computing likeliness. You want
to pick 6 numbers out of 20 available. So for the first pick you
have 20 equally likely outcomes. Let's say you want to pick them in
the order given. You have 2 desirable outcomes. So the likeliness
of getting a desirable outcome on the first pick is (2/20). Next you
want to pick a 3. There are 2 threes and 19 remaining numbers. So
the likeliness of a desirable second pick is (2/19).

The likeliness of getting 6 desirable picks in a row (1, 3, 4, 5, 6,
7 ) is (likeliness 1)*(likeliness 2)*(likeliness 3)* (likeliness 4)*
(likeliness 5)*(likeliness 6)
Or (2/20)*(2/19)*(2/18)*(2/17)*(2/16)*(2/15)
=2^6/(20*19*18*17*16*15)

The likeliness of picking 1, 1, 2, 2, 3, 3 is computed as follows:
Pick first 1, 2 available out of 20 total = 2/20
Pick second 1, 1 available out of 19 total = 1/19
Pick first 2, 2 available out of 18 total = 2/18
Pick second 2, 1 available out of 17 total = 1/17
Etceteras
Or (2/20)*(1/19)*(2/18)*(1/17)*(2/16)*(1/15)
=2^3/(20*19*18*17*16*15)

The likeliness of first ;pick in any order:
(12/20)*(10/19)*(8/18)*(6/17)*(4/16)*(2/15)

The likeliness of the second pick in any order:
(6/20)*(5/19)*(4/18)*(3/17)*(2/16)*(1/15)

As a check,

the likeliness of the second pick in order is 2^8 less than the first
pick in order because in 3 of the picks, there is one available
remaining instead of 2.

the likeliness of the first pick in any order is 6! times the first
pick in the specified order. There are 6! ways of arranging the
order because you have six to choose from for the first choice, and
one fewer choices for each choice that follows.

Regards,
Brian

--- In Math4u@yahoogroups.com, "Kirk Bevins" <hatton02@...> 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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