Download Fundamental Counting Principle

Counting Principles
What you will learn:
• Solve simple counting problems
• Use the Fundamental Counting Principle to
solve counting problems
• Use permutations to solve counting problems
• Use combinations to solve counting problems
Fundamental Counting Principle
• The Fundamental Counting Principle states that if on
e event can occur m ways and a second event can o
ccur n ways, the number of ways the two events can
occur in sequence is m • n.
• Ex. How many different pairs of letters from the
English alphabet are possible?
– There are 2 events in this situation. The first event is the
choice of 1st letter, and the second is the choice of the
second letter.
– 26 x 26 = 676
E2
Telephone numbers in the US currently have 10
digits. The first three are the area code and
the next seven are the local telephone
number. How many different telephone
numbers are possible within each area code?
(Note at this time, a local telephone number
cannot begin with 0 or 1)
E2 Answer
Because the first digit of a local telephone
number cannot be 0 or 1 there are only 8
choices for the first digit. For each of the
other six digits there are 10 choices.
8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000
Permutations
• A permutation is an ordered arrangement of n
different elements.
• A permutation of n different elements is an ordering
of the elements such that one element is first, one is
second, one is third, and so on.
E3
• How many permutations are possible for the
letters A, B, C, D, E, and F?
• Answer
–
–
–
–
–
–
–
–
First position: 6 possible
2nd Position: 5 possible
3rd Position: 4 possible
4th: 3 possible
5th: 2 possible
6: 1 letter remaing
6!=6 x 5 x 4 x 3 x 2 x 1
=720
Number of Permutations of n
Elements
• n • (n—1) • • • • 4 •3 •2 •1=n!
• In other words there are n! different ways that
n elements can be ordered
The formula for the number of permutations of n
elements taken r at a time is
n! .
P

n r
(n  r)!
# in the
collection
# taken
from the
collection
5!  5  4  3  2 1  60
P

P

n r
5 3
2!
2 1
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9
E4
• Eight horses are running a race. In how many
different ways can these horses come in 1st,
2nd, and 3rd? (No ties)
• Answer:
– 1st: 8 possibilities
– 2nd:7 Possibilities
– 3rd: 6 possibilities
– 8 • 7 • 6 = 336
If some of the items are identical, distinguishable
permutations must be used.
In how many distinguishable ways can the letters
STATS be written?
STATS
The T’s
S’s are not distinguishable.
Example continues.
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11
The number of distinguishable permutations of
the n objects is
n!
n1!  n2 !  n3 !  nk !
where n = n1 + n2 + n3 + . . . + nk.
The letters STATS can be written in
5!
 120  30 ways.
2!  2! 1! 4
S’s
T’s
A’s
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12
A combination of n elements taken r at a time is a
subset of the collection of elements where order is
not important.
Using the letters A, B, C, and D, find all the
possible combinations using two of the letters.
{AB}
{AC}
{AD}
{BC}
{BD}
{CD}
This is the same as {BA}.
There are six different
combinations using 2 of the 4
letters.
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13
The formula for the number of combinations of n
elements taken r at a time is
n! .
C

n r
(n  r)!r !
# in the
collection
# taken
from the
collection
4!  4  3  2 1  6
C

C

n r
4 2
2!2! 2 1  2 1
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14
Example 6: How many different ways are there
to choose 6 out of 10 books if the order does
not matter?
3
10!  10  9  8  7  6!  210
C

C

n r
10 6
4!6! 4  3  2 1 6!
There are 210 ways to choose the 6 books.
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15
E7
How many distinguishable ways can the letters
BANANA be written?
Answer:
• There are 6 letters
• 3 are A’s
• 2 are N’s
• 1 is a B
E8
You are forming a 12-member swim team for 10
girls and 15 boys. The team must consist of 5
girls and 7 boys. How many 12-member
teams are possible?
E8
There are ₁₀C₅ ways of choosing 5 girls. There
are ₁₅C₇ ways of choosing 7 boys. By the FCP ,
there are ₁₀C₅ • ₁₅C₇ ways of choosing 5 girls
and 7 boys.
1,621,610