Sample Midterm 2 Key
CSci 2011
(1) (a) False: a1= 1, a2= 3, a3= 7, a4= 15.
(b) False: the strings that both end and begin with ’000’ will be double counted, so you
need to subtract the probability of selecting a bitstring that both ends and begins with
’000’.
(c) True. If a|band b|cthen there are integers pand qso b=pa and c=qb. So b=pqa
and therefore a|c.
(2)(a) The characteristic equation, r2−8r+ 16 = 0, has a double root at 4. So the solution
has the form an=α1(4n) + α2n(4n). Using the values for a0and a1we get a0= 3 = α1, and
a1= 8 = 3(41)+α2(1)(41). So α2=−1 and the solution is an= 3(4n)−n(4n) = (3−n)(4n).
(b) From part (a) the solution to the associated linear homogeneous recurrence is of the
form an=α1(44) + α2n(4n). Since 3 is not a characteristic root, the particular solution is
of the form p0(3n). Substituting this into the nonhomogeneous recurrence gives
p0(3n) = 8p0(3n−1)−16p0(3n−2) + (3n).
Dividing by 3n−2and simplifying gives
p0(9) = 8p0(3) −16p0+ 9
which simplifies further to p0= 9.
Solutions to the linear nonhomogeneous recurrence relation therefore have the form
an=α1(4n) + α2n(4n) + 9(3n).
(3) This is a permutation with indistinguishable objects with 6 objects total: 3 l’s, 2 e’s,
and 1 a. So the number is 6!/(3!2!1!) = 60.
(4)(a) The total number of possibilities is C(10,3). The number of possibilities that have
exactly two of the winning digits is the number of ways of choosing two of the three winning
digits multiplied by the number of ways of choosing any one of the seven nonwinning digits,
so C(3,2)C(7,1). So the probability is C(3,2)C(7,1)/C(10,3).
(b) The number of ways of getting at most one digit correct is the number of ways of getting
no digits correct plus the number of ways of getting exactly one correct. The number of
ways of getting none correct is the number of ways of choosing three of the seven nonwinning
digits, or C(7,3). The number of ways of choosing exactly one is the number of ways of
choosing one of the three winning digits times the number of ways of choosing two of the
seven nonwinning digits, or C(3,1)C(7,2). So the answer is C(7,3)+C(3,1)C(7,2)
C(10,3) .
(5) BASIS STEP: We need to show the fomula holds for n= 1. For n= 1 the sum becomes
4(1)(30) = 4 and the right hand side becomes (2(1) −1)(31) + 1 = 4. This proves the basis
step.
