Ryan Davis
EECS 310 Problem Set 8
1.a) The probability that π(1) = 1 is 1/ |S|. Since π is a random bijection, π(1) is uniformly distributed over the interval [1,
n] with possible discrete outcomes {1,…, n}. There are |S| possible outcomes, each that occurs with equal probability, so
the probability of a single given outcome is 1/ |S|.
b) The probability that f(A) = π(a) is
|S| - π(a)
C
|A|-1
/
|S|
C
|A|
.
The denominator represents the total number of ways to select a subset A of size |A| from S. We choose |A| elements from
a set of |S| elements, the number of ways which to do is simply
|S|
C
|A|
. The numerator is significantly less intuitive. It
gives the number of ways to choose a subset A such that π(a) is the minimum element in A. Since we know that π(a) is the
minimum element of A, A must include |A| - 1 other elements that are greater than π(a). The number of elements that are
greater than π(a) is |S| - π(a), which follows simply from the fact that the original set S has elements {1,…., n}. Thus we
must choose |A| - 1elements from a set of |S| - π(a), given simply by
|S| - π(a)
C
|A|-1.
We give an example to illustrate the correctness of the formula. Consider S = {1, 2, 3, 4, 5}, |A| = 3, and π(a) = 2. The
5
C
3
possibilities for A are
{1, 2, 3}
{1, 2, 4}
{1, 2, 5}
{1, 3, 4}
{1, 3, 5}
{1, 4, 5}
{2, 3, 4}
{2, 3, 5}
{2, 4, 5}
{3, 4, 5}
There are
|S| - π(a)
C
|A|-1
=
5 – 2
C
2
=
3
C
2
= 3 sets in which π(a) = 2 is the lowest number.
c) We can build on the approach described in b to show that the probability that f(A) = f(B) is
(Σ
J=1,…, |S| - 1
|S| - J
C
|A| - 1
*
|S| - J
C
|B| - 1
)
/ (
|S|
C
|A|
*
|S|
C
|B|
).
The denominator gives the total number of ways to choose a subset of cardinality |A| and a subset of cardinality |B|, each
from a set of cardinality |S|. Assume that A and B both have a smallest element J, so f(A) = f(B). A must have A – 1 other
distinct elements, each of which is greater than J. The total number of elements greater than J is |S| - J. Thus the number of
ways to choose a subset A in which J is the smallest element is
|S| - J
C
|A| - 1
. We can perform the same analysis to show that
for the same J, the number of ways to choose a subset A in which J is the smallest element is
|S| - J
C
|B| - 1
. In the numerator,
we sum over all possible smallest elements J to find the total number of ways that we can choose subsets A and B such
that J is the smallest element in both A and B.