PSET 7 Solutions
November 13, 2013
P1. A transitive tournament on n vertices is the same as a ranking of the
vertices. Hence, there are n! of them.
P2. There are n vertices, and each can have at most degree n 1, so each
can have degree 1, 2, . . .
HW 6, Solutions
November 1, 2013
P1. The line notation of a n-permutation is a concatenation of idecompos1
able permutations, each on a block of a partition of [n]. Hence, G(x) = 1F (x) ,
so F (x) = 1 G(x) .
P2. Let A(x) =
n0 an n! . Then, we
PSET 5, Solutions
October 20, 2013
P1. Let A, B and C be the sets of n-permutations containing (1), (2)
and (3) as cycles, respectively. By the inclusion-exclusion principle, We have
|A B C | = |A| + |B | + |C | |A B | |A C | |B C | + |A B C | =
HW 4 Solutions
October 12, 2013
P1. This is equal to the number of permutations of six whose cycle decomposition is given by the product of cycles of length at most 2. Hence, we
have the case of three transpositions, two transpositions, one transpo
September 28, 2013
P1. The same as the number of compositions of 5. Hence, 251 = 16.
P2. This is equal to the number of compositions of 30 into 5 even parts, and
hence equal to the number of compositions of 15 into 5 parts, so 15511 .
P3. We c
September 23, 2013
P1. We need permutations which alternate between even and odd numbers.
We consider two cases, when n is odd and when n is even. When n is even, we
can choose whether the rst term is odd or even, so we obtain 2 n ! n !.
PSET 1, SOLUTIONS
September 15, 2013
P1. SOLUTION ACCEPTABLE ONLY FOR PSET 1: 44n 1 2n 1(
mod 7), so n = 3 works.
SOLUTION EXPECTTED FROM NOW ON (using material covered in the
class): Suppose not, then the set of possible remainders upon dividing b