Math 5705 (Enumerative Combinatorics); Fall 2016
Homework 6 solutions
Problem 1. How many integers between 1 and 1000 are neither perfect squares or perfect
cubes?
Solution. There are 31 squares and 10 cubes between 1 and 1000. The numbers which are
both squares and cubes are precisely the 6-th powers (if a is not a square, then some power in
its prime factorization is odd, and hence a3 is also not a square). There are three 6-th powers
between 1 and 1000. Thus by the Inclusion-Exclusion Principle there are 31 + 10 − 3 = 38
squares or cubes in this range. Hence there are 1000 − 38 = 962 integers which are neither
squares nor cubes in this range.
Problem 2. How many permutations of n have exactly one descent?
Solution. The one-line notation for such a permutation consists of two increasing strings of
integers. To write down such a permutation we need to choose a subset of [n] to be in the
first string, taking into account that that subset cannot be {1, 2, 3, . . . , i} for any 0 6 i 6 n.
Thus there are 2n − (n + 1) choices.
Problem 3. Suppose p, q ∈ Sn . Let p0 = q · p · q −1 . If the cycle notation for p is
(p1 p2 . . . pk1 )(pk1 +1 pk1 +2 . . . pk2 ) . . . (pkl−1 +1 pkl−1 +2 . . . pkl ),
what is the cycle notation for p0 ?
Hint: If p(i) = j, what is p0 (q(i))?
Solution. By definition, if p(i) = j, then p0 (q(i)) = q(p(q −1 (q(i)))) = q(p(i)) = q(j). So the
cycle notation for p0 is
(q(p1 )q(p2 ) . . . q(pk1 ))(q(pk1 +1 )q(pk1 +2 ) . . . q(pk2 )) . . . (q(pkl−1 +1 )q(pkl−1 +2 ) . . . q(pkl )).
Problem 4. Using the Product
Rule
derivatives, prove that for a formal power
0 for formal
F 0 (x)
1
series F (x) with F (0) 6= 0, F (x) = − (F (x))2 . Note that the quotient rule follows from the
product rule and the above reciprocal rule.
Solution. By definition, F (x) ·
Product Rule) yields:
1
F (x)
= 1. Taking the derivative of this identity (using the
1
F (x) ·
+ F (x) ·
F (x)
0
Now solving for the
1
F (x)
0
1
F (x)
0
= 0.
gives the desired result.
Problem 5. Find an explicit formula for the ordinary generating functions
i.e.
P
n>0
of the following sequences (each defined for n > 0):
(1) an = n,
(2) an = n2 ,
(3) an = 3n .
1
an x
n
Solution.
(1)
X
nxn = x
n>0
d 1
x
=
.
dx 1 − x
(1 − x)2
(2)
X
n>0
n 2 xn = x
x
(1 − x)2 − x · 2(1 − x)(−1)
x + x2
d
=
x
·
=
.
dx (1 − x)2
(1 − x)4
(1 − x)3
(3)
X
3n xn =
n>0
2
1
.
1 − 3x