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Emma's Dilemma
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- Words:
- 1011
- Submitted:
- Thu Jul 11 2002
... Emma's Dilemma Arrangements for Emma: emma emam eamm mmae mmea meam mema mame maem amme amem aemm There are 12 possibilities; note that there are 4 total letters and 3 different. What if they were all different like Lucy? Arrangements for Lucy: lucy ucyl cylu ycul luyc ucly cyul yclu lcuy ulcy culy yulc lcyu ulyc cuyl yucl lyuc uycl clyu ylcu lycu uylc cluy yluc There are 24 different possibilities in this arrangement of 4 letters all different. Double the amount as before with Emma's name, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For example there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24. What if there were 4 letters with 2 different? Arrangements for aabb: aabb abab
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