4 letters, no letters repeated. Half or 24 is 12, 4 into 12 is three… Therefore putting this together would make, 4x3=12 x 2=24 = correct answer. This method of calculation is called factorial. 4! = 4x3x2x1 = 24, and there is 24 ways to make Lucy. The reason there is a four in front of the factorial symbol, is because there is for letters in the word. Trying with another word, I’ll see if it works properly and not just with a four letter word.
2 letter word
PJ, JP
There are two arrangements of PJ. Using the equation I worked out above I’ll see if it works. 2! = 2x1= 2 ways, this is correct.
3 letter word
JON, JNO, OJN, ONJ, NOJ, NJO
There are 6 arrangements to this 3 letter word. Using the equation = 3! = 3x2x1 = 6ways. Thus proving my equation right.
So, my equation for part 1, where there are no letters repeated is,
n! =x n= the number of letters in the word. a = number of arrangements.
Using my equation I have worked out this table of results.
Using this equation I can work out that for a 10 letter word there would be 3628800 arrangements for the word.
Part 2
Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, this time using EMMA.
EMMA, AMME, AMEM, EMAM, AEMM, EAMM, MMEA, MMAE, MEMA, MAME, MEAM, MAEM
There are 12 different ways of arranging EMMA. This is a four letter word with two letters repeated. With this in mind, I notice that on the first investigation, I had a four letter word with no letter repeated, now I have a four letter word with two letter repeated. In the first experiment I had 24 different ways and now in this one I have 12. Therefore I notice that 12 is half of 24, which would make sense as two of the letters are repeated in which would half the amount of ways you can rearrange the word. Therefore using part of section 1, I can work out a new equation.
n! / a! = x (number of letters factorial divided by the number of letters repeated equals the number or arrangements.)
Checking that my equation works I will test another name.
3 letter word
DAD, DDA, ADD = 3 ways, 3!/2! = 3 ways, proving my equation to be right.
Now I will try my equation with a 4 letter word with 3 letters repeated.
SMMM, MSMM, MMSM, MMMS
There are 4 different ways of rearranging the letter SMMM. I will now check this with my equation.
4! /3! = 4 ways – proving my equation right again.
5 letter word with 4 letters repeated
CLLLL, LCLLL, LLCLL, LLLCL, LLLLC
There are 5 different ways of rearranging the letters above. Using the equation – 5! / 4! = 5ways, thus making my equation correct.
Part 3
Now I have two equations worked out I have to work out the equation in which will work out the amount of arrangements for any word.
Using a four letter word with two letters repeated this is what I get:
AABB, BABA, BBAA, ABAB, ABBA, BAAB
This is 6 different arrangements. Now all I have to do is link it to the other two parts of this investigation. Doing the first part of the word would be like this, 4! / 2! = 12, then there are two letters left in the word, and that would decrease the amount of ways there are by two, there that would make my equation, 4! / (2!) (2!) = 6.
To test my new equation I have picked a 5 letter word with 2 letters repeated and 3 letters repeated.
FFFLL, FFLFL, FFLLF, FLFLF, FLFFL, FLLFF, LLFFF, LFFFL, LFLFF, LFFLF
There are 10 arrangements of these letters, so using my equation....
5! / (3!) (2!) = 10
Therefore my final equation is
n! / (a!)(b!)(c!)……(z!)
This means number of letters in the word divided by the number of letters repeated.
E.g. A five letter word like tttts; this has 4 t's and 1 s
So: 5! / (1x2x3x4) x (1) = 24 arrangements
A 7 letter word; aabbcdd
7! / (1x2) (1x2) (1x2) (1) = 630 arrangements.
So in conclusion, my final formulae is
n! / (a!)(b!)(c!)…… (z!)
