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1. 1 Emma 2 Emam 3 Eamm 4 Aemm 5 Amem 6 Amme 7 Meam 8 Mema 9 Mame 10 Maem 11 Mmae 12 Mmea There are 12 different arrangements of Emma 2. 1 Lucy 2 Luyc 3 Lyuc 4 Lycu 5 Lcuy 6 Lcyu 7 Culy 8 Cuyl 9 Cylu 10 Cyul 11 Cluy 12 Clyu 13 Yclu 14 Ycul 15 Ylcu 16 Yluc 17 Yucl 18 Yulc 19 Ulcy 20 Ulyc 21 Uylc 22 Uycl 23 Uclu 24 Ucul There are 24 different arrangements of Lucy 3. 1 Jo 2 Oj 1 Joe 2 Jeo 3 Eoj 4 Ejo 5 Oje 6 Oej 1 Ann 2 Nan 3 Nna 1 JJ Number of Letters With Repeated Letters No Repeated Letters JJ Jo 2 1 2 Ann Joe 3 3 6 Emma Lucy 4 12 24 The "no repeated letters" field is equal to the "With repeated letters" field but doubled i.e Lucy = 4 letters = 24 arrangements all you need to do then is to divide it by 2 and you will have the value of "With repeated letters" field and visa versa. There is a quicker solution to this problem. All that you need to do is (If you need a 4 letter word) you would do 1x2x3x4. If you needed 5 letters you would do 1x2x3x4x5 and so on. If you want to find out the value of the repeated letters field you would then just divide it by 2. I predict that a 5 letter word will have 120 combinations because 1x2x3x4x5 =...

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