Chapter 2, Section 6 Elementary Combinatorics

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Chapter 2, Section 6Elementary Combinatorics
? John J Currano, 12/11/2008
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Elementary Combinatorics Counting
Tools Definition. Suppose that A is a finite
set. The cardinality of A is the number of
elements in A, and is denoted A.
Addition Rule for disjoint subsets of a finite
set. If A1, A2, ... , An is a collection of
disjoint subsets of a finite set S, then
Addition Rule for two subsets of a finite set.
If A and B are subsets of a finite set S, then
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Multiplication Rule. Suppose a procedure consists
of m steps, performed sequentially, and that
the 1st step can be performed in n1 ways
the 2nd can be done in n2 ways regardless of how
the 1st was done ? the mth can be
done in nm ways regardless of the previous
choices. Then the number of different ways to
perform the entire procedure is
n1n2nm. Equivalently, the number of ways to
fill m blanks, where the 1st blank can be
filled in n1 ways the 2nd can be filled in
n2 ways regardless of how the 1st was filled
etc. is n1n2nm. Equivalently, the number of
m-tuples where the 1st coordinate can be . . .
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Addition Rule for disjoint subsets of a finite
set, S
Addition Rule for two subsets of a finite set, S
Multiplication Rule A procedure consists of m
steps, performed sequentially, and for each k,
step k can be performed in nk ways regardless of
the choices made on the previous steps. Then the
number of different ways to perform the entire
procedure is n1n2nm.

• Examples. For each experiment, determine the
  number of possible outcomes.
• Equivalently, determine the cardinality of the
  (natural or obvious) sample space for each.
• Toss either a coin or a die, but not both. S
  
• Toss both a coin and a die. S
• Toss a die three times. S

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Addition Rule for disjoint subsets of a finite
set, S
Addition Rule for two subsets of a finite set, S
Multiplication Rule A procedure consists of m
steps, performed sequentially, and for each k,
step k can be performed in nk ways regardless of
the choices made on the previous steps. Then the
number of different ways to perform the entire
procedure is n1n2nm.

• Examples. For each experiment, determine the
  number of possible outcomes.
• Equivalently, determine the cardinality of the
  (natural or obvious) sample space for each.
• Choose one card from a pile consisting of all the
  hearts and face cards from a standard deck of
  cards.
• S
• Arrange three people in a line (1st, 2nd, 3rd).
• S

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Addition Rule for disjoint subsets of a finite
set, S
Addition Rule for two subsets of a finite set, S
Multiplication Rule A procedure consists of m
steps, performed sequentially, and for each k,
step k can be performed in nk ways regardless of
the choices made on the previous steps. Then the
number of different ways to perform the entire
procedure is n1n2nm.

• Examples.
• The die-coin experiment consists of rolling a die
  and then tossing a coin the number of times shown
  on the die. The sequence of coin results (H or T)
  is recorded. How many outcomes are there?
• die shows 1 2 3
  4 5 6
• coin shows (_) or (_,_) or (_,_,_) or
  (_,_,_,_) or (_,_,_,_,_) or (_,_,_,_,_,_)
• coin seqs 21 22 23 24
  25 26
• S 2 4 8
  16 32 64 126

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Elementary Combinatorics Permutations
Consider a set D with n elements. An ordered
sequence of r distinct elements of D is called a
permutation of n objects taken r at a time. In
probability and statistics, a permutation of a
set D of n objects taken r at a time corresponds
to an ordered sample of size r chosen without
replacement from a population D. The number of
permutations of n objects taken r at a time is
denoted by
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Elementary Combinatorics Permutations
To calculate the number of permutations of n
objects taken r at a time, consider filling r
blanks with elements selected from D d1, d2,
d3, . . . , dn the ways to do this is
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Elementary Combinatorics Permutations
Consider a set D with n elements. An ordered
sequence of r distinct elements of D is called a
permutation (of length r). Their number is
Examples. 1. P(n, n) 2. P(n, 1) 3. P(n, 0)
4. The number of permutations of the letters in
the word H I S T O R Y taken 3 at a time is
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Elementary Combinatorics Combinations
Consider a set D with n elements. A subset of r
(distinct) elements of D is called a combination
of n objects taken r at a time. In probability
and statistics, a combination of n objects taken
r at a time from D corresponds to an unordered
sample of size r chosen without replacement from
the population D. The number of such is denoted
by
binomial coefficient
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Elementary Combinatorics Combinations
Consider a set D with n elements. A subset of r
elements of D is called a combination of n
objects taken r at a time.
To calculate the number of combinations, think of
selecting a permutation of r objects from D in
two steps 1. choose a subset of r elements
ways C(n, r) 2. permute the chosen r
elements ways r ! and then use the
multiplication rule P(n, r) C(n, r) ? r ! ?
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Permutations and Combinations
Example. A club has 50 members. 1. In how many
ways can 3 officers be chosen a president,
vice-president, and treasurer, if no one can hold
more than one office? 2. In how many ways can an
executive committee of 3 members be chosen?
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Permutations and Combinations
Example. Ten balanced coins are tossed and the
number of heads is noted. Find the probability
that there are exactly 6 heads. Solution. Even
though we are only interested in the total number
of heads, and even though the coins may be
indistinguishable and may have been tossed all at
once, it is useful to think of this as ten
distinguishable coins tossed in order (e.g., we
could write the numbers from 1 to 10 on them) or
as one coin tossed ten times. This makes the
probability calculation simpler.
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Permutations and Combinations
Restated Example. A balanced coin is tossed 10
times and the sequence of heads and tails is
noted. Find the probability that there are
exactly 6 heads. Solution. Let S be the sample
space of all such sequences and A the subset of
all sequences with exactly 6 heads (and thus 4
tails). We need to calculate P(A) A / S .
We can easily calculate S 210 1024 using
the multiplication rule. How do we
calculate A?
Restated Example. A balanced coin is tossed 10
times and the sequence of heads and tails is
noted. Find the probability that there are
exactly 6 heads. Solution. Let S be the sample
space of all such sequences and A the subset of
all sequences with exactly 6 heads (and thus 4
tails). We need to calculate P(A) A / S .
We can easily calculate S
Restated Example. A balanced coin is tossed 10
times and the sequence of heads and tails is
noted. Find the probability that there are
exactly 6 heads. Solution. Let S be the sample
space of all such sequences and A the subset of
all sequences with exactly 6 heads (and thus 4
tails). We need to calculate P(A) A / S .
We can easily calculate S 210 1024 using
the multiplication rule. How do we
calculate A? Once again we restate the
problem A is the set of all combinations of the
10 coin positions taken 6 at a time that is, to
specify an element of A, a particular sequence
with exactly 6 heads, it is enough to tell which
6 positions contain heads. So A C(10, 6).
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Permutations and Combinations
Restated Example. A coin is tossed 10 times and
the sequence of heads and tails is noted. Find
the probability that there are exactly 6
heads. Solution. Let S be the sample space of all
such sequences and let A be the subset of all
sequences with exactly 6 heads (and thus 4
tails).
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Permutations and Combinations
Example. Ten fair dice are tossed and the number
of ones is noted. Find the probability that there
are exactly 6 ones. What should we do
here? Restate the problem, as in the last example.
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Permutations and Combinations
Restated Example. A fair die is tossed 10 times
and the sequence of faces showing is noted. Find
the probability that there are exactly 6
ones. Solution. Let S be the sample space of all
such sequences and A the subset of all sequences
with exactly 6 ones. We need P(A) A / S .
S 610 using the multiplication rule.
What about A?
Restated Example. A fair die is tossed 10 times
and the sequence of faces showing is noted. Find
the probability that there are exactly 6
ones. Solution. Let S be the sample space of all
such sequences and A the subset of all sequences
with exactly 6 ones. We need P(A) A / S .
S 610 using the multiplication rule.
What about A? A is not simply the set of all
combinations of the 10 die positions taken 6 at a
time to specify a particular sequence with
exactly 6 ones, it is not enough to tell which 6
positions contain ones. We also need to know what
was in the other 4 positions. To see how to
do this, suppose there are ones in positions
1,3,4,5,7,8.
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_ _ _ _ _ _ _ _ _ _ 1 2 3 4 5 6 7
8 9 10
Permutations and Combinations
1
1
1
1
1
1
Any sequence with exactly 6 ones in these
positions can be formed by filling each of the
remaining blanks, in turn, with any number other
than one so each of the 10 6 4 remaining
blanks can be filled in 5 ways. Thus there are 5
? 5 ? 5 ? 5 54 sequences with ones in these six
positions. But we can choose the six positions
for the ones in C(10, 6) ways, so by the
multiplication rule,
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Elementary Combinatorics Partitions of a Set
The binomial coefficient C(n, r) is the number of
subsets of size r of a set S of size n.
When we select a subset A of size r from S, we
partition S into two disjoint subsets of sizes r
and n r, namely A and A. A natural
generalization is a partition of S into a union
of k distinct, pairwise disjoint subsets A1, A2,
..., Ak where k ? 2 and Ai ni for each i
1, 2, ..., k. Of course we must have n1 ???
nk n.
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Elementary Combinatorics Partitions of a Set
Suppose S n and k ? 2. How many different
partitions of S into a union of k distinct,
pairwise disjoint subsets A1, A2, ..., Ak where
Ai ni are there?
The number of such partitions, where the order of
the subsets is important but the order of the
elements within the subsets is not, is the
multinomial coefficient
choose n1 for A1 then choose n2 for A2
then choose n3 for A3 etc.
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Elementary Combinatorics Partitions of a Set
Example. A class has 10 children. Choose 5 to
present a play, 3 to sing, and 2 to play the
piano, with no child chosen more than once. In
how many different ways can this be done?
choose 5 for play then choose 3 to sing
then choose 2 for piano
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Elementary Combinatorics Partitions of a Set
Example. A child has 12 blocks 5 are red, 4 are
green, and 3 are blue. In how many ways can the
blocks be arranged in a line if blocks of a given
color are considered identical? ? ? ? ? ?
? ? ? ? ? ? ? While this may not seem like a
partitioning problem, it can be rephrased as one
partition the set S 1, 2, 3, . . . , 12 of
position numbers or blank labels into three
subsets, R of size 5, G of size 4, and B
of size 3. In the above ex., R1,2,6,9,12,
G3,7,8,10, B4,5,11.
R R G B B R G G R G B R
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Elementary Combinatorics Partitions of a Set
Example. In how many ways can 12 blocks 5 red, 4
green, and 3 blue, be arranged in a line? ? ?
? ? ? ? ? ? ? ? ? ? Restated problem Find the
ways to partition S 1, 2, . . . , 12 into
three subsets R of size 5, G of size 4, and
B of size 3.
R R G B B R G G R G B R
choose 5 for red then 4 for green
then 3 for blue
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Permutations with Repeated Elements
Example. In how many distinguishable ways can the
letters in the word I L L I N O I S be arranged
(permuted)? This is really the same type of
problem as the previous problem. Instead of
arranging a certain number (12) of blocks of
different types (3 colors) with the number of
each color specified, we are arranging a certain
number (8) of symbols (letters) of different
types (5 different letters) with the number of
each type given. So it is again a partitioning of
the set of 8 position numbers, this time into 5
subsets of sizes 3 (I), 2 (L), 1 (N), 1 (O), 1
(S), respectively.
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Permutations with Repeated Elements
Example. In how many distinguishable ways can the
letters in the word I L L I N O I S be
arranged (permuted)? Restated problem Find the
number of ways to partition the setS 1, 2, .
. . , 8 into five subsets I of size 3, L of
size 2, N of size 1, O of size 1, and S of size
1.
choose 3 for I then 2 for L then 1 for N
and finally 1 for S
then 1 for O