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Introduction

Roy Dharmarajan 11N

Maths Coursework

Emma’s Dilemma

Emma’s Dilemma

The name Lucy has got four letters with none of the letters repeated. This means that when I come to choosing a letter I have a choice of four then 3 then 2 then 1 if I write those out and multiply them together I will get the number of arrangements. Which is 4 x 3 x 2 x 1, which also can also be written as 4!

In my coursework I am going to show the different arrangements of Lucy’s name and I am going to find out how many how many arrangements there can be for any name even if the name has more than one letter that is repeated. I will do this by using a formula that I am going to find. Here is a list of all of the arrangements of Lucy’s name:

1. LUCY

2. LUYC

3. LCUY

4. LCYU

5. LYUC

6. LYCU

7. ULYC

8. UCLY

9. ULCY

10. UCYL

11. UYCL

12. UYLC

13. CYLU

14. CYUL

15. CULY

16. CUYL

17. CLUY

18. CLYU

19. YLUC

20. YLCU

21. YCUL

22. YCLU

23. YULC

24. YUCL

There are 24

Middle

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

Conclusion

an="1" rowspan="1">

1

1

---------

---------

---------

--------

--------

2

2

1

---------

--------

--------

--------

3

6

3

1

---------

--------

--------

4

24

12

4

1

--------

--------

5

120

60

20

2

1

--------

6

720

360

120

30

6

1

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.

-  -

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