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letters can be arranged in different orders...

1 letter can b arranged one way round
e.g. = A

2 letters can be arranged two ways round.
e.g. = AB, BA

3 letters can be arranged six ways around.
e.g. = ABC, ACB, BCA, BAC, CBA, CAB

4 letters can be arranged twenty four ways around.
e.g. =  i cant be bothered

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so..   (the the numbers in bold being the amount of letters in the words...)

1=1
2=2x1
3=3x2x1
4=4x3x2x1
5=5x4x3x2x1
6=6x5x4x3x2x1
etc etc...

1=1
2=2
3=6
4=24
5=120
6=720

and so on...

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i understand how it works...

but what i dont undertsand is how i could have this as a formula so i could say

combinations of letter arrangement = .................. (formula)

can anyone shed some light on the subject?

thanks

err, factorials?

1! = 1
2! = 2*1
3! = 3*2*1

so your formula would be x!

quote:

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Originally posted by Ddunk
err, factorials?

1! = 1
2! = 2*1
3! = 3*2*1

so your formula would be x!

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what the fuzz?

EDIT: found: http://www.themathpage.com/aPreCalc/factorial.htm

wow thats confusing

Factorials are basically the number multiplied by every number below it all the way down to 1.

10! = 10*9*8*7*6*5*4*3*2*1

quote:

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Originally posted by M73A
6=6x5x4x3x2x1

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Also equals 6!.

So an equation for your problem would be x! with x being the numbers of letters in the word.

ohhhhhhhhhh that site is starting to make sense now.. thanks

x! god thats simple

ta

wonder if its a word with 2 letters the same

x-amount of letters the same!

quote:

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Originally posted by M73A

quote:

---

Originally posted by Ddunk
err, factorials?

1! = 1
2! = 2*1
3! = 3*2*1

so your formula would be x!

---

what the fuzz?

EDIT: found: http://www.themathpage.com/aPreCalc/factorial.htm

wow thats confusing

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your learning this at 15 ??? we didn't start factorial until 6th form (16/17), even then that was in further maths

quote:

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Originally posted by toddy
your learning this at 15 ??? we didn't start factorial until 6th form (16/17), even then that was in further maths

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lol... erm, it dont seem to hard untill i read further down the page...
dont think we're going into to much depth

I don't know if you're doing this, but if there are repeated letters there is a formula for it.

eg. AAABB

n = total letters = 5
x = no. of As      = 3
y = no. of Bs      = 2

number of permutations formula = n! divided by (x! * y!) where x and y are the number of times the repeated letters occur.

n! = 5*4*3*2*1 = 120
x! = 3*2*1 = 6
y! = 2*1  = 2

120/(6*2) = 120/12 = 10 permutations

Don't know if you need that and sorry if i didn't explain it well enough, but it might come in handy.

Just to explain a bit more, say you had the word AAAABBBCCD

n = 10 (total letters)
w = 4 (number of As)
x = 3 (number of Bs)
y = 2 (number of Cs)
z = 1 (number of Ds)

formula = n! / (w!*x!*y!*z!)

             = 10! / 4!*3!*2!*1!

             = 3628800 / 288

             = 12600 permutations

Thanks Monster.rat. . . that is what i needed!

it was on the site i found but your wording was earier to understand

quote:

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Originally posted by M73A
Thanks Monster.rat. . . that is what i needed!

it was on the site i found but your wording was earier to understand

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No problem

Glad i didn't type all that up for nothing. I did some maths coursework investigation on that last year and its still on my computer, so just looked at that.