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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 Spring '09
 pts, Natural number, positive integers, arbitrary integer n=k

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