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Emma’s Dilemma

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Introduction

Emma's Dilemma My name is Andrew Donald MacKay; this name has 18 letters and two spaces in it. * 4 A's * 3 D's * 2 N's * 1 R * 1 E * 1 W * 1 O * 1 L * 1 M * 1 C * 1 K * 1 Y * 2 Spaces How many different ways can I write the letters of my name? To investigate this problem I started with shorter names, e.g.) Lucy Jo oj 2 Tim tmi itm imt mit mti 6 Lucy luyc lyuc lycu lcyu lcuy ulcy ulyc uylc uycl ucly ucyl cluy clyu cylu cyul culy cuyl yluc ylcu yulc yucl yclu ycul 24 Jason jasno jaosn jaons janso janos jsaon jsano jsoan jsona jsnao jsnoa joasn joans josan josna jonas jonsa jnaso jnaos jnsao jnsoa jnoas jnosa ajson ajsno ajosn ajons ajnso ajnos asjon asjno asojn asonj asnjo asnoj aojan aojna aoajn aoanj aonja aonaj anjao anjoa anajo anaoj anoja anoaj sjaon sjano sjoan sjona sjnao sjnoa sajon sajno saojn saonj sanaj sanoj sojan sojna soajn soanj sonja sonaj snjao snjoa snajo snaoj snoja snoaj ojsan ojsna ojasn ojans ojnsa ojnas osjan osjna osajn osanj osnaj osnaj oajsn oajns oasjn oasnj oanjs oansj onjsa onjas onsja onsaj onajs onasj njsoa ...read more.

Middle

Bob bbo obb 3 - 3! = 6 Aa 1 - 2! = 2 Every time a letter is repeated twice, the number of combinations is half. This expression needs to be expanded to include what happens when a letter is repeated more than twice. Bobby bboby bboyb bbboy bbbyo bbybo bbyob bobyb boybb byobb bybob bybbo obbby obybb obbyb oybbb yobbb ybobb ybbob ybbbo 20 - 5! = 120 The fraction of reduction (the fraction of the number of combinations when letters are repeated over number possible when all letters are different) should remain constant when number of letters are changed but number repeated stays the same. Ebneee ebenee ebeene ebeeen eebnee eebene eebeen eeebne eeeben eeeebn eenbee eenebe eeneeb eeenbe eeeneb eeeenb enbeee enebee eneebe eneeeb nbeeee nebeee neebee neeebe neeeeb bneeee beneee beenee beeene beeeen 30 - 6! = 720 These are 3 fractions of reduction. Number Of Times Letter Used Fraction Of Reduction 2 1/2 3 1/6 4 1/24 Once again they are following the factorial sequence. Each number of combinations is divided by the factorial of the number of repeated letters. So I can adjust my original expression of u = n! ...read more.

Conclusion

- all adding up to 2 x 3!. This is equal to 12; this is the same number that according to my expression is removed by division. This proves my expression to be accurate. The fraction of reduction remains the same for any size of word because the combinations always repeat themselves in the same patterns. If my tree diagram was for a 5-letter word, 1 of the 5 main branches would be completely removed and 1 would be completely kept. The other 3 would follow the same pattern as above. A 3-letter word would follow exactly the same pattern as 1 of my main branches in the diagram above. As the number of letters in the word goes up, so the patterns of repeated combinations goes up in proportion. What about spaces? To work out the different combinations of series of words or sentences, the placing of spaces also becomes important. Spaces can be counted as a 27th letter. So to calculate the number of combinations of Andrew Donald MacKay... ____________18!____________ 4! (a) x 3! (d) x 2! (n) x 2! (space) = 11115232128000 This expression can be used to calculate the number of different combinations of any combination of letters or numbers or anything else. Page 1 Andrew MacKay 11SM ...read more.

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