Letter permutations

Ahoyhoy · Joined: 05 Aug 2007 Posts: 66

Given that 1 letter can be arranged 1 way:

A

2 letters can be arranged 2 ways

AB
BA

3 letters can be arranged 6 ways

ABC
ACB
BAC
BCA
CAB
CBA

4 letters can be arranged 24 ways

ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
BACD
BADC
BCAD
BCDA
BDAC
BDCA
CABD
CADB
CBAD
CBDA
CDAB
CDBA
DABC
DACB
DBAC
DBCA
DCAB
DCBA

I know that the number of permutations can be found with an factorial
calculation:

so 2 letters = 1 x 2 = 2 permutations

and 3 letters = 1 x 2 x 3 = 6 permutations

and 4 letters = 1 x 2 x 3 x 4 = 24 permutations

therefore 5 letters = 1 x 2 x 3 x 4 x 5 = 120 permutations

Is there a quick way of working out the number of permutations for, say, 25
letters without having to do the full factorial calculation?

Archived from group: uk>education>maths

spike1 · Joined: 05 Aug 2007 Posts: 149

Ahoyhoy did eloquently scribble:
> Is there a quick way of working out the number of permutations for, say, 25
> letters without having to do the full factorial calculation?

Well... there's always the factorial button on most modern scientific
calculators.

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Frederick Williams · Joined: 05 Aug 2007 Posts: 20

Ahoyhoy wrote:

> Is there a quick way of working out the number of permutations for, say, 25
> letters without having to do the full factorial calculation?

Are you seeking an approximation to n! for large n? If so, look up
Sterling's approximation.

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Stan Brown · Joined: 05 Aug 2007 Posts: 233

Sat, 01 Dec 2007 10:54:04 GMT from Ahoyhoy :
> Is there a quick way of working out the number of permutations for, say, 25
> letters without having to do the full factorial calculation?

http://www.google.com/search?
q=approximation+factorial&sourceid=mozilla-search&start=0&start=0
&ie=utf-8&oe=utf-8

That's a Google search for "approximation factorial" (no quotes).

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