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Emma's dilemma.

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Introduction

Joe Grix 10H         of         Maths Coursework

For this investigation, I am going to investigate the number of different arrangments of letters in a word. An example of arrangments of a word can be JAM or MAJ etc, the same word but how many times the word can be mixed up to make different words or arrangments of the letters.

Here is a key to show the meanings to different letters and symbols

Letter

Meaning

A

Number of arrangments

L

Number of letters

S

Number of letters the same

+

Plus

x

Multiply

/

Divide

-

Subtract

=

Equals

The number of arrangments for words with the number of letters from 1 – 5, and no pairs of letters so there is only 1 same.

  1. J                Number of arrangments = 1
  1. JO
  2. OJ                Number of arrangments = 2
  1. JOE
  2. JEO
  3. OJE
  4. OEJ
  5. EJO
  6. EOJ        Number of arrangments = 6
  1. LUCY
  2. LUYC
  3. LCUY
  4. LCYU
  5. LYCU
  6. LYUC
  7. ULYC
  8. ULUC
  9. UCYL
  10. UCLY
  11. UYLC
  12. UYCL
  1. CLYU
  2. CLUY
  3. CULY
  4. CUYL
  5. CYLU
  6. CYUL
  7. YLUC
  8. YLCU
  9. YULC
  10. YUCL
  11. YCLU
  12. YCUL

Number of arrangments = 24

  1. JAMES
  2. JAMSE
  3. JAEMS
  4. JAESM
  5. JASME
  6. JASEM
  7. JSAEM
  8. JSAME
  9. JSEAM
  10. JSEMA
  11. JSMEA
  12. JSMAE
  1. JMAES
  2. JMASE
  3. JMESA
  4. JMEAS
  5. JMSEA
  6. JMSAE
  7. JEASM
  8. JEAMS
  9. JEMAS
  10. JEMSA
  11. JESMA
  12. JESAM

24 arrangments so far.  x 5  because there are 24 arrangments starting with ‘J’, so there will be 24 ways for each other letter ‘A,M,E,S’, and there are 5 letters including ‘J’.

5 x 24 = 120

Number of arrangments = 120

Table to show all of the number of arrangments with different number of letters, with only 1 letter the same.

Number of letters

Number of arrangments

1

1

2

2

3

6

4

24

5

120

...read more.

Middle

The number of arrangments for words with the number of letters from 1 – 5, and 1 pair of letters so there is 2 same.

  1. JJ        Number of arrangments = 1
  1. MUM
  2. MMU
  3. UMM                Number of arrangments = 3
  1. EMMA
  2. EMAM
  3. EAMM
  4. AMME
  5. AEMM
  6. AMEM
  7. MMEA
  8. MMAE
  9. MEMA
  10. MEAM
  11. MAEM
  12. MAME        Number of arrangments = 12

Number of arrangments         = 24 + 36

  1. HARRY
  2. HARYR
  3. HAYRR
  4. HRRAY
  5. HRRYA
  6. HRAYR
  7. HRARY
  8. HRYRA
  9. HRYAR
  10. HYARR
  11. HYRAR
  12. HYRRA

Here there are 12 arrangments starting with the letter ‘H’, as there are 3 single letters, and 1 pair, then 12 x 3 = 36, the number of arrangments starting with a single letter.

  1. RRHAY 1
  2. RRHYA 2
  3. RRAHY 3
  4. RRAYH 4
  5. RRYAH 5
  6. RRYHA 6

123 4

Here there are 6 arrangments starting with there letter ‘R’, and second with the other letter ‘R’ there are 4 letters altogether that can be the second letter after the first ‘R’. so if there’s 6 arrangments with 1 letter in second after the first letter, then there will be            4 x 6 = 24 arrangements, because there are 4 other letters that can go second.

        = 60


The number of arrangments for words with the number of letters from 1 – 5, and 1  3 letters the same, so 3 same.

  1. JJJ                Number of arrangments = 1
  1. NUNN
  2. NNUN
  3. NNNU
  4. UNNN        Number of arrangments = 4
  1. SASSY
  2. SASYS
  3. SAYSS
  4. SYSSA
  5. SYSAS
  6. SYASS
  7. SSSAY
  8. SSSYA
  9. SSYAS
  10. SSYSA
  11. SSASY
  12. SSAYS
  1. ASSSY
  2. ASSYS
  3. ASYSS
  4. AYSSS
  5. YSSSA
  6. YSSAS
  7. YSASS
  8. YASSS

Number of arrangments = 20

...read more.

Conclusion

As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.

As before, the original formula:

L! = the number of letters in the word

 S! = the number of letters the same

From this I have come up with a new formula  The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.

For the above example:

A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

So : 1x2x3x4 _        24

       (1x2) x (1x2)         4     = 6 different arrangements

A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)

So: 1x2x3x4x5

      (1x2x3x4) x (1)

           = 120

               24     = 5 different arrangements

A five letter words like abcde; this has 1 of each letter (no letters the same)

So : 1x2x3x4

   (1x1x1x1x1x1)

          = 24                      = 24 different arrangements

A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).

So : 1x2x3x4x5

       (1x2x3) x (1x2)

= 120

    12    = 10 different arrangements

This shows that my formula works:

L!   = the number of letters in the word

       X!Y! = the number of repeated letters the same

...read more.

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