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Emma’s Dilemma  

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John Palfrey, 5UL Emma's Dilemma 1) Rearrangement Of EMMA If we represent the letters in Emma's name as numbers, we get an arrangement like this: E M M A 1 2 2 3 We can then use this system to find all of the other potential arrangements: 1 2 2 3 1 2 3 2 1 3 2 2 2 1 2 3 2 1 3 2 2 2 1 3 2 2 3 1 2 3 1 2 2 3 2 1 3 1 2 2 3 2 1 2 3 2 2 1 As you can see, there are 12 combinations. 2) Rearrangement Of LUCY If we represent the letters in Lucy's name as numbers, we get an arrangement like this: L U C Y 1 2 3 4 We can then use this system to find all of the other potential arrangements: 1 2 3 4 2 1 3 4 1 2 4 3 2 1 4 3 1 3 2 4 2 3 1 4 1 3 4 2 2 3 4 1 1 4 2 3 2 4 1 3 1 4 3 2 2 4 3 1 3 1 2 4 4 1 2 3 3 1 4 2 4 1 3 2 3 2 1 4 4 2 1 3 3 2 4 1 4 2 3 1 3 4 1 2 4 3 1 2 3 4 2 1 4 3 2 1 As you can see, there are only 24 potential arrangements; twice as many as when there is one repetition. 3) Other Rearrangements If we use the name MAY to investigate the number of potential arrangements when there are only three factors, we get these arrangements: M A Y 1 2 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 We only get six arrangements, a quarter of the amount when we use four factors. If we use the word MOO to investigate the number of potential arrangements when there are only three factors, we get these arrangements: M O O 1 2 2 1 2 2 2 1 2 2 2 1 We only...

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