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Emma’s Dilemma

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Thomas Collyer 10Lr                MATHS

27/04/07                EMMA’S DILEMMA

Page  of



To investigate the patterns caused by the permutations of letters in words of different lengths and to investigate the possibility of predicting the number of permutations. To discover a formula that can be applied to all words.

Question 1.

12 different permutations can be made by the name EMMA:

  1. EMMA
  2. EMAM
  3. EAMM
  4. MMEA
  5. MMAE
  6. MEMA
  7. MEAM
  8. MAEM
  9. MAME
  10. AMEM
  11. AMME
  12. AEMM

Question 2.

...read more.


  1. ELICV
  2. ELIVC
  3. ELCVI
  4. ELCIV
  5. EICVL
  6. EICLV
  7. EILVC
  8. EILCV
  9. EIVCL
  10. EIVLC
  11. EVCIL
  12. EVCIL
  13. EVLIC
  14. EVLCI
  15. EVILC


4) Looking at mathematics behind answers

The word Lucy has 4 letters l,u,c and y. If we know how many letters (no repeats) are in a word we can work out the number or permutations by multiplying 1 x 2 x 3 x 4 and so on until we reach the correct number of digits in a word (e.g. a 10 letter word would be 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10). The mathematical symbol for this is n! and stands for number factorial, this shows number of letters! . After working this out we can quickly and easily work out the number of permutations made by a word with repeats – this id done by dividing our previous answer by 2. For 2 repeats we divide the answer again by 2.

...read more.


! q! ...  

(In words: n factorial over p factorial times q factorial and so on)


How many permutations can be formed from the letters, taken 5 at a time, of the word DADDY?

Solution  (has 3 letters that are the same – D)

5! / 3! = 20

  • The 5! Shows the word has five letters in it including repeats.
  • The 3! Shows the word has 3 letters that are the same (d)


Find the number of arrangements of all the letters in the word MARMALADE.


9! / 2! x 3! = 30240

  • The 9! Shows the word has 9 letters altogether including repeats.
  • The 2! Shows the word had 2 letters the same (m).
  • The 3! Shows the word has 3 more letters that are the same (a).

This formula can be used conclusively for other problems.

In how many ways can 3 apples, 2 oranges, 4 pears and one banana be given to 10 children if each child receives a piece of fruit?


10! / 3! x 2! x 4! = 12600

...read more.

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