Math 184A Second Exam Solutions 2 March 2005
1. 0
1
2
3
4
5
You can use theorems, but it is probably easiest to color 0 (xways), then color
1, 3 and 5 (x−1 ways each since each must differ from 0) and finally color 2 and 4
(x−2 ways each since each must differ from two differently colored vertices—0 and 1
or 0 and 3). This gives x(x−1)3(x−2)2.
2. (a) Use induction. True for n= 1 and n= 2 by inspection. For n > 2, the left (resp.
right) tree has leaves of length 1 + (n−1) (resp. 2 + (n−2)).
(b) It is evident that no leaf has adjacent B’s since we either prepend A or BA. To
see that every such sequence arises, use induction. It’s true for lengths 1 and 2
by inspection. For n > 2, the sequence must begin either A or BA and then is
followed by any sequence with no adjacent B’s. By the induction hypothesis for
n−1 and nall such sequences occur in S∗(n−1) and s∗(n−2).
3. (a) We have (1 −x−2x2)A(x) = x. Equating coefficients of xn, we have
an−an−1−2an−2equals 3 if n= 1 and 0 otherwise.
Treating n≤1 as initial conditions, this gives us
a0= 0, a1= 3, an=an−1+ 2an−2for n≥2.
Alternatively, introducing cn= 0 except that c1= 3 gives us
an=an−1+ 2an−2+cnfor all n.
(As usual, anis assumed zero for negative n.)
(b) Since 1 −x−2x2= (1 −2x)(1 + x), partial fractions gives us
A(x) = b
1−2x+c
1 + x=Xb(2x)n+Xc(−x)n=X(b2n+c(−1)n)xn.
Thus an=b2n+c(−1)n. You can find b= 1 and c=−1 either by the usual
partial fraction route or by solving the two equations 0 = a0=b+cand 3 = 2b−c.
4. Each such tree is either a single vertex (xy) OR a root (x) joined to two trees (T(x, y)2)
OR a root joined to four trees, etc. Thus we have
T(x, y) = xy +X
k≥1
keven
x(T(x, y))k=xy +x(T(x, y))2
1−(T(x, y))2.