Emma's Dilemma

Now I am going to investigate the number of arrangements in Emma’s name:

1. EAMM     

2. EMAM     

3. EMMA    

4. MEAM   

5. MMEA   

6. MMAE   

7. MAEM

8. MAME

9. AMME                        

10. AEMM

11. AMEM

12. MEMA 

To work this out without writing all the combinations down I will have to do 4 x 3 x 2 x 1 then divide the answer by 2. The reason that I am dividing it by 2 is because there are 2 letters that are repeated so I will have to divide it by 2 to get the right answer. I get half the results for Emma because there are 2 m’s so it is like just having 1 m. If I had just 1 m then it would be EMA then I would of got 12 different combinations for this name and even if I have 2 M’s it would make no difference because both M’s are the same there is nothing special or different to the M’s.

This is a one-letter word with no letters repeated

1.A

This name only has one combination so if I do 1! It will tell me that the amount of combinations is one.

This is a two-letter word with no letters repeated

1. AL
2. LA

As you can see there are 2 different combinations so if I do 1x 2 or 2! Then the answer will be 2 this shows that my formulae have worked for a two-letter word.

This is a three-letter word with no letters repeated

1. TOM

2. TMO

3. OMT

4. OTM

5. MOT

6. MTO

As you can see there are 6 different combinations that shows that my formulae works with any name with no repeated letters. This formulae works for me if none of the letters are repeated. I have tried the formulae with different amounts of letters with no letter repeated, and I kept on getting the correct answer. This means that this formulae works with any name with no letters repeated.

Now I have to find a formula so that I can work out how many different combinations there are in any name with any amount of letters repeated.

This is a 2-letter name with 1 letter repeated 2 times:

1. AA

This name has 2 letters repeated and there is only 1 combination for this name.

Now I am going to do a 3-letter name with 1 letters repeated 2 times and see what happens.

1. ABB
2. BAB
3. BBA

This name has 3 different combinations.

Now I want to see what happen if I have a name with a 4-letter name with 1 letters repeated 2 times.

1. ABCC

2. ACBC

3. ACCB

4. CABC

5. CACB

6. CCBA

7. CCAB

8. CBCA

9. CBAC

10. BACC

11. BCAC

12. BCCA

Now this is a 4-letter word and there are 12 different combinations for it.

Now I have figured out a formulae to find out how many different combinations there are without writing out a list of combination and it is n! / r this means the number of letters in the name factorial divided by the number of letters repeated. Now to work out the amount of combinations in the name above I will use my formulae.

4 x 3 x 2 x 1 = 24

24/2 =12

And I have reached the correct answer this shows that my formulae is correct.

Now I am going to do a name that has 5 letters with 1 letter repeated 3 times.

1. AAABC
2. AAACB
3. AABCA
4. AABAC
5. AACBA
6. AACAB
7. ABACA
8. ABCAA
9. ABAAC
10. ACABA
11. ACAAB
12. ACBAA
13. BCAAA
14. BACAA
15. BAACA
16. BAAAC
17. CBAAA
18. CABAA
19. CAABA

20.CAAAB

Now a 5-letter word with 1 letter repeated 3 times will give me 20 different combinations. I have now got a formula for this and it is n!/r! this means number of letters factorise divided by the number of repeated letters. I am dividing by number of repeated letters factorised because that way gives me the results that I need. I have tested this formula on all the names I done before as well and I have got all of the answers correct using this formula. This means that this formula will work on any name with any letter repeated more than once.

Now here is a table to show all the different results that I have got from all different amounts of repeated amounts of letters with different amounts of letters:

Amount of Letters Repeated

I will now investigate the different combinations of X’s and Y’s                            

Here are all of the different combinations of XXXYYY:

1. XXXYYY
2. XXYXYY
3. XXYYXY
4. XXYYYX
5. XYXXYY
6. XYXYXY
7. XYXYYX
8. XYYXYX
9. XYYYXX
10. XYYXXY
11. YYYXXX
12. YYXYXX
13. YYXXYX
14. YYXYXX
15. YXYXYX
16. YXYXXY
17. YXXYXY
18. YXXXYY
19. YXYYXX
20. YXXYYX

There are 20 different combinations of XXXYYY, I have a formulae which I can use to get the amount of combinations for XXXYYY without writing a list of combinations and it is: n!/x!/y!. So if I do (6!/3!/3!= 20) I will get the correct answer 20 this shows that my formulae works.

Now I will give you a general formula, which can be used to find any the answer to any combination, and it is: (n!/a!/b!/c!)

And if there are more letters, which are repeated, then just add another divide and divide it by the amount of times that letter is repeated.