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Mathematics GCSE EMMA'S DILEMMA 1. No. of arrangements Arrangements 1 EMMA 2 MEAM 3 AEMM 4 MMEA 5 MMAE 6 AMEM 7 EMAM 8 AEMM 9 MAEM 10 AMME 11 MAME 12 MEMA 2. No. of arrangements Arrangements 1 LUCY 2 LYCU 3 LYUC 4 LUYC 5 LCYU 6 LCUY 7 UCYL 8 UYCL 9 ULCY 10 ULYC 11 UYLC 12 UCLY 13 CYLU 14 CYUL 15 CLUY 16 CLYU 17 CULY 18 CUYL 19 YLUC 20 YLCU 21 YUCL 22 YULC 23 YCLU 24 YCUL 3. No. of arrangements No. of letters Position(n) 1 A 1 2 AB 2 6 ABC 3 24 ABCD 4 120 ABCDE 5 720 ABCDEF 6 5040 ABCDEFG 7 40320 ABCDEFGH 8 362880 ABCDEFGHI 9 3628800 ABCDEFGHIJ 10 I have completed the table by using a method I found after finding a pattern in the first few results. For a four-letter word (all different letters) there are 24 arrangements as I found in Question 2 (LUCY). There are six combinations beginning with each letter, with this knowledge I think that a five-letter word (all different letters) would have 120 arrangements. To prove this answer/back it up I have found another pattern in the table. 4 letter word 1x2x3x4=24 (No. of arrangements) (all different) 5 letter word 1x2x3x4x5=120 (all different) 6 letter word 1x2x3x4x5x6=720 (all different) I then finished off the table up to a 10-letter word (all different letters). Next I decided to do a further investigation, to investigate the number of different arrangements of words where 2 letters are the same. Take the word aabb (4 letters, 2 different). aabb has six different arrangements: 1....

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