Wheels and Permutations

[Moses]

Hello to all

How many different settings for one set of seven numbers, for example;
01,02,03,04,05,06,07
Would I be correct to say that the above sequence has 2401 different ways of appearance?

Thanks


Mo

[Frank]

Moses Wrote:Hello to all

How many different settings for one set of seven numbers, for example;
01,02,03,04,05,06,07
Would I be correct to say that the above sequence has 2401 different ways of appearance?

Thanks


Mo

Well If you are meaning all the possible different orders of 7 balls, these are permutations. I think its 5040 - the formula in Excel is =PERMUT(7,7)
..Frank

[Moses]

Hello Frank and thanks

I would have thought that each set has seven wheels and each number has seven positions which is equal 49 and this calculations will be for each number then it would be 49X49 = 2401

I am not in to Excel but I am trying to find out how you will work it, however I have no doubt that your calculation is correct but I like to know what I am missing out!


Regards,

Mo

[Frank]

Moses Wrote:Hello Frank and thanks

I would have thought that each set has seven wheels and each number has seven positions which is equal 49 and this calculations will be for each number then it would be 49X49 = 2401

I am not in to Excel but I am trying to find out how you will work it, however I have no doubt that your calculation is correct but I like to know what I am missing out!


Regards,

Mo

Yes, Moses I understand your difficulty with this and you ar not alone - permutations are more difficult to get your head around than combinations because order matters.
Lets take a simpler example which we can quickly work out manually and see why the above logic doesn't work. Suppose we had just 3 numbers , lets call them A,B,C . How many different orders can you find for them ?
A B C
A C B
B A C
B C A
C A B
C B A

I think that's it, and =permut(3,3) gives you an answer of 6 in Excel.
Using the same logic you had for 7 numbers, you might say:-
" I would have thought that each set has three wheels and each number has three positions which is equal 9 and this calculations will be for each number then it would be 9X9 = 81 .( Which we know is wrong.)

We haven't always had Excel and there is a longhand way of doing the calculation described here by DR Math :-
PERMUTATIONS & COMBINATIONS The top one third of the page applies to combinations.
If you modify his example to calculate the permutations of 7 balls within a population of 7, you get:-

7_P_7 = (Factorial 7 )Divided by [factorial (7-7)]
= (7x6x5x4x3x2x1 ) divided by Factorial 0
=5040 /1 ... (factorial zero is 1 , by the way)
7_P_7=5040
So that agrees with the answer given by Excel. Its to do with factorials just as lottery odds calculations involve them.

Regards, Frank

[Moses]

Hello Frank and thanks again

One reason that I am so keen to learn the breakdown is to workout the probabilities for these

18/04/2007, 30,41,17,08,11,32,47
11/04/2007, 08,42,35,21,11,32,47

08/03/2003, 32,05,08,06,33,41,15
24/10/1998, 15,41,26,06,08,23,49 this is in exact reverse order

17/02/2006,04,23,38,24,26,04,02 Euro
04/06/2005,04,13,07,24,26,04---- Thunder

There far too many situations like above had happened in lotto or other draws!

I will look at the link however to see if I can learn something form it.

Thanks

Mo