exam2sol - COT 3100 Exam #2 Solutions 11/7/02 Name: _...

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COT 3100 Exam #2 Solutions 11/7/02 Name: ____________ Lecturer: Arup Guha TA: _______________ (Note: You will have 75 minutes for this exam. Make sure to read AND follow all the directions. If you need extra room for your work, put it on the last page of the exam and CLEARLY number what problem’s work you are continuing.)
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1) Permutations: You must explain your reasoning in order to get full credit on these questions. a) (3 pts) How many permutations are there of the word HALLOWEEN? Using the formula for permutations with repetition, since we have 2 Ls and 2 Es, we have 9!/(2!2!) = 90720 b) (3 pts) How many of the permutations in part a start and end with the same letter? Two groups of mutually exclusive permuations to count: 1) Those that start and end with L, and 2)those that start and end with E. Each of these groups will have an equal number of permutations because to get all the permutations in group 2 from the ones in group one, we could just exchange the Ls and the Es. The number of permutations in group 1 = 7!/2! since you have 7 letters left to permute with one letter repeated twice. Thus, the final answer is 2(7!/2!) = 7! = 5040 c) (4 pts) How many of the permutations in part a do not contain any adjacent vowels? __ C __ C __ C __ C __ C __ Let the 5 consonants be places where the Cs appear the vowels can be placed in any of the blank slots, this ensures that none of them are adjacent. Since the placement of the vowels is independent to the ordering of the consonants, we can multiply the number of orderings of the consonants by the total number of possible placements for the vowels for our final answer. The total number of orderings of the consonants
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exam2sol - COT 3100 Exam #2 Solutions 11/7/02 Name: _...

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