Emma’s Dilemma.

1. JOE
2. JEO

So you then multiply 2 by 3 to give you the answer 6.

1. JOE
2. JEO
3. OJE
4. OEJ
5. EJO
6. EOJ

You can now look at a five letter word, I am going to look at JIMMY.

1. JIMMY
2. JIMYM
3. JIYMM
4. JMMYI
5. JMMIY
6. JMIYM
7. JMYIM
8. JMYMI
9. JMIMY
10. JYMMI
11. JYMIM
12. JYIMM
13. MMYJI
14. MMJYI
15. MMIYJ
16. MMYIJ
17. MMJIY
18. MMIJY
19. MJIMY
20. MJMIY
21. MJYMI
22. MJIYM
23. MJMYI
24. MJYIM
25. MIJMY
26. MIJYM
27. MIYJM
28. MIYMJ
29. MIMJY
30. MIMYJ
31. MYMJI
32. MYMIJ
33. MYIJM

As you can see there are 60 ways of writing JIMMY, so if we look at my first rule again, we can see that there are 12 ways of writing JIMMY beginning with the letter J. So we come up with the sum 12 X 5 = 60, so my rule works again. But what if we started with the letter M instead of J, we can see that there are 24 ways of writing JIMMY beginning with M, so the sum that we get is 24 X 5 = 120. This is the answer given to us if we marked the double letters 1 and 2 for example in this case M1 and M2. So to give us the answer of 60 we need to divide this answer (120) by the number of letters that are the same, in this case 2, we then get the sum 120 ÷ 2 = 60.

I will now look at the name PIPPA, notice that it have 3 letters the same.

All of the combinations beginning with P:

1. PPPIA
2. PPPAI
3. PPAIP
4. PPIAP
5. PPIPA
6. PPAPI
7. PAIPP
8. PIAPP
9. PAPPI
10. PIPPA
11. PIPAP
12. PAPIP

As you can see there are 12 ways of writing PIPPA beginning with the letter P so using my rule I can predict that there will be 20 ways of writing PIPPA, I know this because, the sum works out as:

12 (number of ways of writing PIPPA beginning with P) X 5 (number of letters in the word)

3 (number of letters the same in the word).

OR MORE SIMPLY:

12 X 5

     

As you can see there are, indeed 20 ways of writing PIPPA.

1. PPPIA
2. PPPAI
3. PPAIP
4. PPIAP
5. PPIPA
6. PPAPI
7. PAIPP
8. PIAPP
9. PAPPI
10. PIPPA
11. PIPAP
12. PAPIP
13. IPPAP
14. IAPPP
15. IPAPP
16. IPPPA
17. AIPPP
18. APIPP
19. APPIP
20. APPPI

I am now going to look at a number of different words all with letters repeated. I predict that there will be 10 ways of writing XXXYY. I know this because there are six ways of writing it with X at the beginning so the equation that I get is 6 x 5  =  10.

                                                                     3

1. XXXYY        1. YYXXX
2. XYYXX        2. YXYXX
3. XXYYX        3. YXXYX
4. XYXXY        4. YXXXY
5. XYXYX        
6. XXYXY        

3 letter word with 2 letters repeated.

1. XXY
2. XYX
3. YXX

4 letter word with 3 repeated.

1. XXXY
2. XXYX
3. XYXX
4. YXXX

5 letter word with 4 repeated.

1. XXXXY
2. XXXYX
3. XXYXX
4. XYXXX
5. YXXXX

6 letter word with 5 repeated.

1. XXXXXY
2. XXXXYX
3. XXXYXX
4. XXYXXX
5. XYXXXX
6. YXXXXX

So once we have worked out how many ways there are of writing a word all beginning with a certain letter then we can work out the total number of combinations of that word even if it has letters repeated.

I have now put my results into a table so that I can look for any patterns that I have.

We can see from my results that 3 X 2 = 6

And that 4 X 6 = 24 and that 5 X 24 = 120

6 X 120 = 720 and 7 X 720 = 5040

        

This shows us that there is another way of working out the number of different combinations. It is called factorial, written 5!. What this does is 5 X 4 X 3 X 2 X 1 = 120. This is the total combinations of letters for a five letter word that has no letters repeated.

If we look at all of my results for words that have got letters repeated then we will get a bigger picture.

                                 Half of 5040        A Third 2520         A Quarter of 840

We can simply see from this table that each time that you add a letter, to be repeated, then the number of possible combinations of that word is reduced by a factor of firstly 2 then by 3 and then by 5.

All of the combinations that are in Bold Italic I have worked out using my rule.

I will now try to prove my rule.

Lets take the combination of letters XXYYZZ, I will now predict and prove both of my rules with this combination.

First of all:

1. XXYYZZ
2. XXZZYY
3. XXYZYZ
4. XZYZXY
5. XZXYYZ
6. XZYYZX
7. XZYZYX                        
8. XZZXYY                        X 6 = 90                
9. XZZYXY
10. XYZZYX                                        
11. XYZYXZ
12. XYXZYZ
13. XYZZYX
14. XYZYZX
15. XYZZXY

Then second:

                        6! (number of letters in the word)        

                    2! X 2! X 2! (number of letters the same)

          =

                                  720                

                             2 X 2 X 2

          =

                        720                

                              8

           =

                90

Well both of the rules seem to work so, my final equation for my second rule turn out to be

           N!                

       R! X R!

N = number of letters in the word.

R = number of letters repeated in the word.