CSCE 235
Handout
22: Proofs in Combinatorics
March 26, 2004
Permutations
with Some Identical Objects
The number of different
permutations of n objects, where there are 
indistinguishable
objects of type 1, 
indistinguishable
objects of type 2, …, and 
indistinguishable
objects of type k, is
Proof is very
elegant. To determine the number of
permutations, first note that the 
objects of type one
can be placed among the n positions in 
ways, leaving 
positions free. Then, the objects of type two can be placed
in 
ways, leaving 
positions free. And so
on, until the last stage 
objects of type k can
be placed in 
. Hence, by the
produce rule, the total number of different permutations is:
.
Repetitions
with Identical Objects
This addresses the question: how
many ways are there can we put r identical objects into n boxes
and multiple objects can go into the same box? Suppose we want to put 10
identical marbles into 3 boxes. One way
to put the marbles into the boxes is to put three marbles into box 1, five into
box 2, and two into box 3. This
situation can be represented by the 12-digist string 000100000100, which has
strings of three, five, and two 0s separated by two 1s. There is a one-to-one correspondence between
ways of putting the marbles into the box and 12-digit strings consisting of two
1s and ten 0s. The string 001000000100
would represent the situation of two marbles in box 1, six in box 2, and two in
box 3. The string 100000010000 would
indicate that there are no marbles in box 1, six in box 2, and four in box
3. To count the number of ways of
putting ten marbles in three boxes is, therefore, just to count the number of
12-digit 0-1strings which contain precisely two 1s. This latter number is easy to find! There are 
such ways. Therefore the number of ways to put 10
identical marbles into 3 boxes, any number to a box, is 
. This is in the form
of 
, the number of ways to put r identical marbles into n
boxes.
Derangements
A derangement of n distinct symbols
which have some natural order is a permutation in which no symbol is in its
correct position. The number of
derangement of n symbols is denoted 
.
Consider 
, the number of derangements of 1, 2, 3, 4. Let 
be the set of
permutations of 1, 2, 3, 4 in which the number 1 is in the first (and correct)
position. Let 
be the set of permutations
of 1, 2, 3, 4 in which the number 2 is in the second (and correct)
position. Define 
and 
similarly. The set of permutations in which at least
one of the four members is left in its correct position is then 
and the complement of
this set consists precisely of the derangements of 1, 2, 3, 4. Since there are 
permutations
altogether, the number of derangements of four symbols is
.
How many elements are in the union of
? By the Principle of
Inclusion-Exclusion, we have
Now 
(1 is I nthe first
position, numbers 2, 3, and 4 go to any of the next three positions). Similarly, 
for any i. The set 
contains those
permutations of 1, 2, 3, 4 in which 1 and 2 are in the correct positions. There are just two such permutations, 1234
an 1243. Thus, 
for each of the 
terms in the second
sum on the right of the above equation.
Similar reasoning shows that each of the 
terms 
equals 1, as does the
last term. Therefore, 
. Hence,
.
To find the general formula for 
, for general n, we mimic this calculation of 
.
Proofs
are excerpted from (Goodaire and Parmenter 2002).