Download Multiplication Principle and Addition Principle

Multiplication Principle and
Addition Principle
1. Multiplication Principle: Suppose a task is
accomplished by n steps and each step requires a
choice from a number of available choices. Let
these numbers be A1, A2, . . . An
Then, the total number of ways to accomplish
this task is
A,
A1 x A2 x A3 x . . . . An.
1
2. Addition Principle: Suppose a task is
accomplished by choosing an object from the
union of disjoint sets with cardinalities B1, B2,
B3, . .. Bn. Then, the total number of ways to
accomplish this task is
B1+ B2+ B3 + . . . . . Bn.
Set Theoretic Descriptions
of the two principles
• The cardinality of the Cartesian product of n sets A1, A2,
. . An is given by
n (A1 x A2 x . .. An) = n(A1). n(A2) . . .n(An).
• The cardinality of the disjoint union of n sets
B1, B2, . . . . Bn is given by
n (B1 U B2 U . . . Bn)= n(B1) + n(B2)+ . . . + n(Bn)
Examples
• On a table, there is a pile of 10 apples and
another pile of 15 pears.
1. How many choices do you have if you
are instructed to pick an apple and a pear?
2. How many choices do you have if you
are instructed to pick an apple or a pear?
Mistakes you must avoid
• Suppose there are 4 doctors and 5 chess
players in a room with 7 people. If you are
instructed to pick a doctor or a chess
player in this room (and slap the face) how
many choices do you have?
Is your answer 4+5 = 9 ? Then, review your knowledge of
addition principle.
Read the problems carefully.
In a room, there are 7 people. Everyone in that room is either a chess player
or a doctor. 4 of them are doctors and 5 of them are chess players. In how
many ways can one accomplish each of the following tasks?
1. Task is to get into the room, pick a chess player and play a chess with.
Then, pick a doctor and ask for an opinion about your asthma symptom.
2. Task is to pick a team of two people, consisting of one chess player and one
doctor.
• Addition Principle is frequently applied in a
form of `subtraction’ principle.
n(A \ B) = n(A) – n(B)
Example: In a basket of 30 Easter eggs, there are 7 green eggs. Your
task is to pick an egg that is not green. How many choices do you
have?
The answer:
30-7= 23.
23 choices.
Another Example: (See the Venn diagram of chess players and
doctors.) We will approach the second question in the previous
slide in the following way. Initially, we write a name of a chess player
and (next to it) write a name of a doctor. Consider this writing as a
tentative list of 2-team members. There are 5x4=20 possible ways of
writing a pair of names in this way. But, exactly 2 of them are a pair
of names of a same person (which is not permitted). So, the answer
to the second question ( of forming a team of chess_player-doctor
pair) is 20-2=18.
Multiplication Principle is frequently applied in the
form of Division Principle.
• Division Principle:
If A=B x C (cartesion product of B and C), then
n(B) = n(A) / n(C).
EXAMPLE 1: Gregor Samsa goes to walk every morning wearing a hat
and carrying a cane. He has 2 hats, which are identical except that
one is grey and the other is brown. He has 3 canes. He is color
blind. In how many ways, can he choose a hat and a cane to go out
for a walk based on his discerning ability?
Answer: If he were able to tell brown from grey, there would be
2 x 3= 6 ways. But, now that all the hats are identical to
him
from his view, there are 6 / 2 = 3 ways.
Division Principle re-stated:
Suppose a task A consists of accomplishing a task B followed by a task C.
Then, the number of ways to accomplish task B is given by the number of
ways to accomplish the task A divided by the number of ways to accomplish
task C.
EXAMPLE 2. How many different words can be formed by rearranging the
letters in the word “ELEMENT”?
Answer:
7! / 3!
( Explain. . . . .)
Permutation
• Informally, a permutation on a set of n elements is an ordering of
these elements.
• Formally, a permutation on a set of n elements is a one-to-one
correspondence between the set and itself.
• Number of permutations on a set of n elements
• Generalized Permutation: Let A be a set of n elements and r<
n. Any one-to-one function from the set
{1,2,3, . .. . r} into the set A is called a generalized permutation of
the k elements on the set A.
• Number of generalized permutation of r elements on the set of n
elements is denoted by nPr.
•
nPr x
(n-r)! = n!
by multiplication principle.
Combination
• Let r<n. An r-combination on the set of n
elements is a subset with cardinality r.
• nCr denotes the number of all the rcombinations on the set of n elements.
• nPr and nCr are related by the following
equation
nPr = nCr x r! (illustrated below)
• Picking a generalized permutation,
a one-to-one function from {1,2, . . r} into
The set A of n elements is equivalent to
STEP1: Picking a subset of A that has r elements.
STEP 2: Picking a permutation on these r elements.
Therefore,
nPr= nCr
x
r!
Summary
• n! = number of permutations on n objects.
n!
• nPr = number of r-permutations on n objects =
(n  r )!
•
nCr = number of r-sets that can be formed from n objects
Pr
=n
/ r! =
n!
(n  r )! r !
Partition
• Let A be a set of n elements. If, for some subsets A1, A2,
. . . , Ak that are pairwise disjoint,
A= A1 U A2 U . . . U Ak
then we say {A1, A2, . . . Ak} is a partition of A.
Ordered Partition
• Let a1+a2+ . . . ak= n and a1< a2< . . ak
( < here means less than or equal to)
The sequence of subsets
A1, A2, . . . . Ak with cardinalities
a1, a2, . . . . . ak
is called an ordered partition if these sets
are mutually disjoint and their union is the
set A.
Number of ordered partitions
•
•
How many ordered partitions of a given type does a set of n element have?
Experiment: Consider a set of 7 elements. How many ordered partitions of type
(2,2,3) are there? The construction of a partition of a given
type can be considered as a multi-step task. The number of
these steps is 1 less than the number of subsets that form the
partition. In this particular experiment, 2 steps.
STEP 1: Pick a subset of 2 elements (from 7 elements)
STEP 2: Pick a subset of 2 elements (from the remaining 5 elements)
Now, we employ the multiplication principle:
7
C 2 x 5 C 2 = 7! 5!  7!
2!5! 2!3! 2!2!3!
Counting Ordered Partitions of a
Given Type
Let
a1+a2+ . . . ak = n.
The number of
type (a1, a2, . . . . ak) partitions of a set of n
elements is given by
n!
a1!a2!a3!   ak !
Application (of partition counting)
• How many different words can be formed
by rearranging the letters in the word
“ MISSISSIPI” ?
Discussion: Compare this problem with the
following problem: `` How many ordered
partitions of the type (1,1,4,4) are there for
a set of 10 elements?”.
How are these two problems related?
• Given a 10 element set, imagine these
elements are doors along a hall way of a
hotel. Imagine your duty is to write
letter M on 1 door
letter P on 1 door
letter S on 4 doors
letter I on 4 doors
This job is equivalent to forming a word by an
arrangement of letters in “MISSISSIPI”.
This job is also equivalent to forming a type (1,1,4,4)
ordered partition of the 10 element set.