Emma's Dilemma.
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Introduction
Emma’s Dilemma  
We are investigating the number of different arrangements of letters. Firstly we arrange Emma’s name…
There are 12 different arrangements for the name Emma. Note that there are four letters in total and three different letters. Next I arranged Lucy…
There are 24 different possibilities in this arrangement of the name Lucy with four letters that are all different. Twice as many as before in Emma’s name which had four letters but three different letters. In the name Lucy I noticed that there were six possibilities beginning with a different letter each time. For example in the name Lucy there were six arrangements starting with L six starting with U six starting with C and six starting with Y. Therefore I found that 6 times the amount of letters (in this case 4) equaled the amount of different arrangements for the name. If we take aabb as a further example for the amount of arrangements, what would happen? Arrangements for aabb…
There are six different arrangements for the letters aabb as the table shows. From Emma with two different letters to Lucy with all different letters I have found a pattern for the amount of arrangements (6, 12, 24). It is easier for me to convey more difficult arrangements through tables so I will do a table with more letters in the it, and try to look for an explanation for the different amount of arrangements. I will start with a fiveletter word. I have previously found that with a four letter word with all different letters there are 24 separate arrangements and that there were six ways of starting the arrangements i.e. start with an L first then a U and so on and so on. For the sake of argument I have added a Q to the front of Lucy to make a five letters word with no repeated letters. I have predicted that qlucy will have 120 different arrangements. I came to this conclusion because in Lucy there 24 arrangements and if this was divided by four (the number of letters in Lucy) it would equal six, the amount of different possibilities there are for each separate letter. Therefore in the name qlucy there will be 24 arrangements starting with L 24 arrangements starting with U and so on and so on. So 120 divided by five (the number of letters) equals 24. To help prove that with a fivelettered word with no repeats there are 24 possibilities for each letter, I will draw a table using qlucy as my example…

Middle
Ycul
Luyc
Ucly
Cyul
Yluc
Yluc
Lycu
Uycl
Culy
Yucl
Lcuy
Ulyc
Cylu
Yclu
Lcyu
Uylc
Clyu
Ylcu
Note that the arrangements start with L then on to U then on to C and lastly on to Y. Each time a letter is used it cannot be used again. So therefore, every time the letter is used it is “usedup”. This leaves the word with one less to be arranged with i.e. for lucy 4x3x2x1. In this case the L is used first followed by the U then the C and then the Y. This is how I decided upon using factorial to express this.
I put in my results on my calculator to see if it works on the calculator and the figures were correct. Now that I have proof that factorial works on my calculator I can rely on it to provide me with the arrangements for words with no repeats. To save time I put in 6! And it gave me 720, this number make sense because 720 divided by 6 (the number of letters in the word) gave my 120. This is the amount of arrangements for a fiveletter word and the numbers continue to fall in this pattern.
Total Amount Of Letters Number Of Arrangements
1 1
2 2
3 6
4 24
5 120
6 720
I have now explained the pattern and formula for n amount of letters in a name with no repeats. Now I know this formula I need to know what the formula is for a name with repeated letters. Take Emma as an example; the name has four letters in it but two are repeated (MM). If I used the formula for no repeats such as N! I found that for a four lettered word it gave 24 combinations. When I arranged Emma I only found 12 combinations. This suggests that there is more than N! to the formula for repeated letters in a name. To allow you and I to understand what is happening more easily I will replace letters with X’s and Y’s. These letters stand for any letters and any amounts of repeats. I will start with xxyy.
XXYY is a fourletter word with two different letters.
xxyy xyxy yxxy
xyyx yxyx yyxx
Now I will have five letters and have xxxyy.
xxxyy xxyxy xxyyx xyxyx xyxxy
xyyxx yyxxx yxxxy yxyxx yxxyx
Now I will have five lettered word xxxxy.
xxxxy xxxyx xxyxx
xyxxx yxxxx
There are only five different arrangements for this set of letters.
If we go back to xyyz we can see that 4! equals 24. I have noticed that if you divide this by two (the number of repeated words) it give you the correct amount of combinations. Also in a fivelettered word there are 120 arrangements and 24 starting with a different letter of the word. From this I can see that there is going to be a divide in the formula. If we again take xyyz we can use this formula N! / 2… this gives us the correct amount of arrangements for the name xyyz. I have tried this formula on many other names in the table bellow ;
Name  Amount Of Letters  Arrangements  Repeats  Divide By? 
xyyz  4  12  2  2 
Lxyyz  5  60  2  2 
Qlxyyz  6  360  2  2 
Wqlxyyz  7  2520  2  2 
Conclusion
N! / R!
Proof;
Name  Amount Of Letters In Name  Repeats  Arrangements  Correct (✓ / X) 
Emma  4  2  12  ✓ 
Lucy  4  0  24  ✓ 
Anna  4  4  1  X 
Hanna  5  4  5  X 
As the table shows the formula works with no repeats and with one repeat. Any more than one repeat and the formula doesn’t work.
What Would Happen If I Were To Split The Repeats Into Separate Equations?
Example;
Anna
4! / (2 x 2)! =
1
Therefore this formula is incorrect as the right amount of arrangements for this name is 24.
What, if instead of the repeated letters are times together and factorialised it is reversed?
The repeated letters are factorialised and timesed together.
So for sake of argument instead of having to R’s call one X. The X is now in the equation because if we have more than one equation we need to times by these repeated letters. The formula is now;
N! / R! x X!
Try this on the previous name.
Name  Amount Of Letters In Name  Repeats  Arrangements  Correct (✓ / X) 
Emma  4  2  12  ✓ 
Lucy  4  0  24  ✓ 
Anna  4  4  24  ✓ 
Hanna  5  4  120  ✓ 
The formula works!!!
The formula is correct so now I will do further calculations to A) show how to use the formula and to B) further my investigation.
Calculations.
HANNAH
N! / R! X X!
6! / 2! X 2! X 2!
= 1440 combinations
CHARLIE
N! / R! X X!
7! / 0! X 0!*
= 5040 combinations
Note: even if there are no repeats the formula still works because you just add in 0’s and this doesn’t affect the formula and still arrives at the right amount of combinations.
The formula can work with any amount of letters or numbers. I’ll use X’s and Y’s again to prove this.
Total Amount Of Letters  Number Of X’s  Numbers Of Y’s  Total Amount Of Arrangements 
3  2  1  3 
4  2  2  6 
5  2  3  10 
6  2  4  15 
3  3  0  1 
4  3  1  4 
5  3  2  10 
6  3  3  20 
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