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27th June 2001 Tom Pountain 10A Emma's Dilemma I investigated the number of different arrangements of four letters with no repetitions. 1)ABCD 2)ABDC 3)ADBC 4)ADCB 5)ACBD 6)ACDB 7)BACD 8)BCAD 9)BDCA 10)BADC 11)BDAC 12)CDBA 13)CBAD 15)CDAB 16)CADB 17)CBDA 18)DABC 19)DBCA 20)DCAB 21)DACB 22)DBAC 23)DCBA 24)DABC 25)BCDA I have found 24 different arrangements of these letters and this result is confirmed in the tree diagram. Secondly, I have investigated the number of different arrangements of four letters with one letter repeated twice. 1)ACBB 2)ABCB 3)BABC 4)BACB 5)BBAC 6)BBCA 7)BCAB 8)BCBA 9)CBBA 10)CABB 11)CBAB 12)ABBC I have found 12 different arrangements of the letters and this result is confirmed in the tree diagram. From these two investigations, I have worked out a method that can be used for further work: Firstly, with ABCD you rotate the last two letters, then you get ABDC. Then, with ABCD you must then rotate the last three letters and try the possibilities of ADBC, ADCB, ACDB, ACBD. Because the letter 'B' has been the first number of last three letters...

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