Mathematial Problem - Printable Version
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Mathematial Problem by M73A on 01-03-2006 at 11:27 PM
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
====================================================
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...
=====================================================
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
RE: Mathematial Problem by ddunk on 01-03-2006 at 11:29 PM
err, factorials?
1! = 1
2! = 2*1
3! = 3*2*1
so your formula would be x!
RE: Mathematial Problem by M73A on 01-03-2006 at 11:30 PM
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
RE: Mathematial Problem by ddunk on 01-03-2006 at 11:34 PM
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: Originally posted by M73A
6=6x5x4x3x2x1
Also equals 6!.
So an equation for your problem would be x! with x being the numbers of letters in the word.
RE: Mathematial Problem by M73A on 01-03-2006 at 11:37 PM
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!
RE: RE: Mathematial Problem by toddy on 01-03-2006 at 11:38 PM
quote: 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
your learning this at 15 ??? we didn't start factorial until 6th form (16/17), even then that was in further maths
RE: Mathematial Problem by M73A on 01-03-2006 at 11:41 PM
quote: 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
lol... erm, it dont seem to hard untill i read further down the page...
dont think we're going into to much depth
RE: Mathematial Problem by Zephyr on 01-03-2006 at 11:55 PM
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
RE: Mathematial Problem by M73A on 01-04-2006 at 12:10 AM
Thanks Monster.rat. . . that is what i needed!
it was on the site i found but your wording was earier to understand
RE: Mathematial Problem by Zephyr on 01-04-2006 at 12:21 AM
quote: 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
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.
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