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Unformatted text preview: Answers to Homework 14 February 2008 Section 15 4. How many di/erent anagrams (including nonsensical words) can be made from FACETIOUSLY if we require that all six vowels must remain in alphabetical order (but not necessarily contiguous with each other) . I guess he considers Y to be vowel. I wonder about the usage &contiguous with each other I would have said simply &contiguous. Anyway, you can rst choose the six positions for the vowels you have no choice on the order of the vowels within those six positions. The number of ways to pick the six positions is & 11 6 = 462 Then you have to place the ve consonants. Their order is not prede termined so you have 5! = 120 ways to put them in their ve positions, which are determined by the placement of the vowels. So the answer is 462 & 120 = 55 ; 440 . Alternatively, you can rst place the ve consonants, the rst in any of 11 positions, the second in any of the remaining 10 , and so on for a total of 11 & 10 & 9 & 8 & 7 = 55 ; 440 ways. Then you place the vowels, but you have no choice! This seems like an easier way to calculate this number. 9. You wish to make a necklace with 20 di/erent beads. In how many di/erent ways can you do this. The usual assumption on making a necklace is that the linear arrangements abcdefghijklmnopqrst rstabcdefghijklmnopq tsrqponmlkjihgfedcba all give the same necklace. Of course its possible to construct necklaces where this is not the case. However, my guess is that the author has this in mind. The number of linear arrangements of 20 beads is 20! . This overcounts the number of necklaces by a factor of 20 , because linear arrangements like the rst two displayed above give the same necklace, and also by a factor of 2 because linear arrangements like rst and third give the same necklace. So the answer is 20!...
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 Spring '08
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