permutation

[claybizz]

how do you perm 5 from 43

[gen2]

To perm N numbers from a pool of y objects

There are y ways of selecting the first number leaving y-1 objects
There are y-1 ways of selecting the second number leaving y-2 objects
There are y-2 ways of selecting the third number leaving y-3 objects
and so on until . . .
There are y-(N-1) ways of selecting the Nth number leaving y-N objects

The total number of permutations is found by multiplying these together

= y * (y-1) * (y-2) * . . . . * (y-(y-1))

This can also be written as y! / (y-N)!
(the ! means factorial)

[gen2]

Oops! Sorry. Correction:

I mistyped the formula 3 lines from the end which should have read

= y * (y-1) * (y-2) * . . . . * (y-(N-1))

[Peter Pedant]

and if you do it in figures

43 ways to choose the first

42 to choose the second (43x42) in all

41 ways to chose the third 43x42x41

go on like this for two more lines and you get an expression 43x42....

which is also 43! / 38!

[c00ky83]

Divide by 5! = 120 to avoid repetitions! e.g. 12345 = 54321, etc.

[claybizz]

Many thanks your prompt replies. This has prompted me to find out the difference between Perms and Combinations. Not bad for a 70 year old duffer.

[gen2]

Im glad you got it.
Permutations are where the order DOES matter.
Combinations are where the order does NOT matter.

c00ky83 tells you above how to go from permutations to combinations. You divide by the factorial of the number of items in the set - in this case 5!

[claybizz]

Thanks again gen. I've done a few examples and got it for sure now. I have an HNC (from Roman times) in Telecomms engineering but haven't done any maths for ages. This site with people like you will be a great help in keeping the old brain alive.