Combinations of LettersDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/