Wheels and Permutations

[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