score780 wrote:
Zarrolou wrote:
score780 wrote:
How many different arrangements can be made with the letters in the word signing? (I there an algebraic way to solve it?)
signing= 7 letters
Tot arrangements = 7! = 5040
You can choose 7 letters for the first position, 6 letters for the second, ... (with no repeat)
So you obtain 7*6*5*4*3*2*1=7!
Not really. The answer is actually 7!/(2!*2!*2!) because of the repeated letters. I am just wondering if there is a way to explain this in a step by step algebraic way using permutation law.
Hi score780, let me try and explain how this works in the general case and then go about using a formula.
Let's take a different 7 letter word with no repeating letters: History. Each letter is different so if you try and change the order you get a different answer. There are 7x6x5x4x3x2x1 or 7! (5040) ways to rearrange these letters).
If we take a 7 letter word with only one repeating letter: Wishing. Each letter can be identified by a separate number, but rearranging the i's will yield a new permutation that is indistinguashible from the first iteration. I.e. w
1sh
1ng and w
ish
ing. This means that we have 7!/2! (or 2520) ways of writing out the word wishing.
If we take the 7 letter word signing with repeating i's, n's and g's, we get a total of 7! ways to rearrange the letters divided by 2! for identical i's, 2! for identical n's and 2! for identical g's, yielding a total of 7!/2!*2!*2! or 630 ways of writing it out.
In the general case we take \(N!/A!B!...\) where N is the total number of letters and A is the number of times element A is repeated, B is the number of times element B is repeated, etc.
Hope this explanation makes sense. You can just use the formula if you want but it's always interesting to try and understand why the formulae hold.
Thanks!
-Ron