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Combinations of Letters
Date: 10/20/98 at 03:04:55
From: Anthony
Subject: General formula for permutations and combinations
To Dr. Math:
I have been set a piece of work involving finding a general rule for
the different number of arrangements that can be made out of people's
names, such as EMMA (12 combinations) and LUCY (24).
I have come up with this formula:
Number of letters!
---------------------------
Number of different letters
e.g.:
6!
Hannah = --------
2!x2!x2! (because it has 2 h's, 2 n's, and 2 a's)
10!
Commitment = -----
3!x2! (because it has 3 m's and 2 t's)
What I need your help on is how to put this formula into a general
term. I have found the top line (N!) but I am having difficulty with
the bottom one. Could you please show me how you deduce the formula?
Yours in anticipation,
Anthony
Date: 10/20/98 at 10:48:55 From: Doctor Rob Subject: Re: General formula for permutations and combinations Count the frequencies of the letters that appear. In COMMITMENT, C -> 1 O -> 1 M -> 3 I -> 1 T -> 2 N -> 1 E -> 1 Note that the sum of these numbers must be the total number of letters: 1 + 1 + 3 + 1 + 2 + 1 + 1 = 10 Then the numerator is the factorial of the sum, and the denominator is the product of the factorials of the counts: 10!/(1!*1!*3!*1!*2!*1!*1!) = 10!/(3!*2!) = 302400 since 1! = 1. (If you like, you can throw in the other letters, like Q and Z, with frequency counts 0. They don't change the sum since 10 + 0 = 10, nor the product of the factorials of the counts, since 0! = 1.) - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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