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Emma's Dilemma.

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Minesh Patel 11B        Maths Coursework        Mr Murray


Emma and Lucy are playing with combinations of the letters of their names. One arrangement of Lucy is YLUC. Another is LYCU.

Part 1 – The different combinations of the letters of Lucy’s name.

Below are all the different combinations of the name Lucy.

  1. LUCY
  2. LUYC
  3. LCYU
  4. LYCU
  5. LCUY
  6. LYUC
  7. UCYL
  8. UCLY
  9. UYLC
  10. UYCL
  11. ULYC
  12. ULCY
  13. CLUY
  14. CLYU
  15. CULY
  16. CUYL
  17. CYUL
  18. CYLU
  19. YLUC
  20. YLCU
  21. YUCL
  22. YULC
  23. YCLU
  24. YCUL

The name Lucy has 24 different combinations. This means that all names with 4 different letters will have a total of 24 different combinations.  Looking at other, similar, names with 4 different letters such as Mark, we can see this.

Below are all the different combinations for the name Mark.

  1. MARK
  2. MAKR
  3. MRKA
  4. MRAK
  5. MKAR
  6. MKRA
  7. ARKM
  8. ARMK
  9. AKRM
  10. AKMR
  11. AMKR
  12. AMRK
  13. RKMA
  14. RKAM
  15. RMKA
  16. RMAK
  17. RAKM
  18. RAMK
  19. KRAM
  20. KRMA
  21. KMAR
  22. KMRA
  23. KARM
  24. KAMR

As you can see the name MARK also has 24 different combinations.


The number of combinations for the name Lucy can be worked out in the following ways.

  1. no. of letters + 2  x  no. of letters      [4+2 x 4 = 24]
  2. no. of letters factorial                        [4! = 1x2x3x4 = 24]

To see if the above methods work, we can test them on the three (different) letter name Ian.

Below are the different combinations of the name Ian.

  1. IAN
  2. INA
  3. ANI
  4. AIN
  5. NIA
  6. NAI

We can see that the name Ian has 6 possible combinations. Now I will test the methods mentioned previously.

No. of Letters + 2 x No. of Letters = No. of Combinations

[        3            + 2 x         3           =            15               ]

...read more.













Part 2– The different combinations of Emma’s name

Below are all the different combinations of the name Emma.

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

As you can see, the name Emma has only 12 different combinations in contrast to the 24 the name Lucy has. This is because, Emma has two letters in it which are the same, so the number of combinations for a four (different) letter word, would be halved, to give the results.

Calculations _                                                                        

We can use the following method to finds the different number of combinations for a four letter word with two letters the same.                

No. of Letters Factorial ÷ No. of letters the same = No. of Combinations                  4 !               ÷               2 !                =             12

The above method is correct as it gives the correct number of combinations. This is because an arrangement of 4 different letters provides 24 different combinations, however in this case, 2 of the letters are the same, and as the combinations cannot be repeated, the number of letters factorial must be divided, by the number of repeated letters factorial. I can guarantee the above method is correct by taking the arrangement of letters, ABBBC.

...read more.


÷ (n1 x n2 x etc) [X being the number of letters, and n being the number of repeated letters.].

Linking Formulas

        The General Formula is No. of Combinations = X ! ÷ (n1 ! x n2 ! x etc.)

        This formula can be applied to any set of letters as the formula works on the basis of primarily finding out the number of combinations for multiple sets of repeated letters. The n1 or n2 can be replaced with 1 in order to find out the number of combinations for one repeated set of letters. This is because there is no extra set of repeated letters, so X! eventually is divided by n!. You can replace the n1 and n2 by 1 in order to find out the number of combinations for a set of completely different letters. This is because if n is equal to 1 then it is the same as having no repeated letters so the formula can be simplified to X!.

        There is no exception, as the formula allows you to work out the number of combinations for completely different letters, a set of repeated letters and multiple sets of repeated letters.

...read more.

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