5.3 Combinations and Permutations
Specific Curriculum Outcomes
A6
G7
G8
develop an understanding of factorial notation and apply it to calculating
permutations and combinations
distinguish between situations that involve combinations and permutations
develop and apply formulae to evaluate permutations and combinations
Assumed Prior Knowledge
q
q
q
q
calculating probability
independent and dependent events
conditional probability (3205 only)
simulation
Investigation 10
Combinations and Permutations
Purpose:
to define and distinguish between combinations and permutations
Procedure:
Most bankcards can be used to access an account by entering a 4-digit personal
identification number (PIN). However knowing these four digits is not enough to
access the account. The digits have to be in the correct order.
Consider Leida’s account, where four digits used in opening it are 1, 3, 7, and 9.
Leida knows the first digit used is 1 but cannot remember the order of the
remaining digits. We can arrange these remaining digits in six different ways.
379
397
739
793
937
973
The order is important, so we must consider each of these as a different
arrangement. There are six possible arrangements to consider and only one of
these arrangements will access the account. These types of arrangements are
called PERMUTATIONS.
Permutation: an arrangement of a set of objects
If the order were not important, then in this case it would suggest that all six of
these arrangements would access the account.
Combination: a selection from a group of objects without regard to order
1
Focus F
Factorial Notation
Consider how many ways are there of arranging 6 different books side by side on a
shelf.
We have to calculate the product of 6 x 5 x 4 x 3 x 2 x 1.
In mathematics we denote this product by 6! (called “6 factorial”).
6! = 720
In general n! = n(n – 1)(n – 2)(n – 3) … (3)(2)(1).
10! 3628800
=
= 720
7!
5040
OR
1.
Express as single factorials.
a)
5x4x3x2x1
b)
8 x 7 x 6 x 5!
c)
(n + 3)(n + 2)(n + 1)n(n - 1)… x 3 x 2 x 1
2.
Find the value of the following:
13!
 26!  5! 
a)
b)

 
10!3!
 23!  9! 
3.
Simplify:
a)
4.
n!
(n − 3)!
b)
10! 10 × 9 × 8 × 7!
=
= 10 × 9 × 8 = 720
7!
7!
(n + 2)!
n!
Express as a quotient of factorials:
a)
10 x 9 x 8 x 7
b)
21 x 20 x 19
c)
n!
n( n − 1)
c)
(n + 2)(n + 1) n
c)
( n + 1)!
= 20(n − 1)
(n − 2)!
3205 only
5.
Solve
a)
(n + 1)!
=7
n!
b)
( n + 1)! = 6(n − 1)!
Answers
1. a) 5! b) 8! c) (n + 3)!; 2. a) 286 b) 524/63; 3. a) n 3 - 3n2 + 2n b) n 2 + 3n + 2 c) (n – 2)!; 4. a) 10!/6! b) 21!/18! c) (n + 2)!/(n - 1)!;
5. a) 6 b) 2 c) 4
2
Investigation 11
Writing an Expression for Permutations
The written driver’s test consists of 25 questions randomly selected from a pool of
100 questions. This allows for a large number of different possible tests. When a
test is created, there are 100 possibilities for the first question, 99 for the
second, and so on. The number of possible tests is therefore 100 x 99 x 98 x … x
77 x 76.
Procedure:
A:
Use the Fundamental Counting Principle to write the expression for the
number of different driver’s tests.
B:
Rewrite the expression from step A as a ratio of factorials.
C:
Rewrite the denominator using 100, the number of items in the test bank,
and 25, the number of items on each test.
D:
Let n be the number of items in the test bank and r be the number of items
n!
on each test. Explain why the expression
gives the number of
(n − r )!
permutations, taken r items at a time from a total on n items.
E:
Use the formula from step D to find the number of possible driver’s tests.
3
Investigation Question
15.
Evaluate 6 P4 using the expression from step D, without using a calculator.
Permutations with Repetitions
The number of permutations of n objects, where a are the same of one kind, b are
the same of another kind, and c are the same of yet another kind, can be
represented by the expression:
n!
a!b!c!
Question:
Solution:
Find the number of permutations of the letters of the word
MATHEMATICAL.
There are 2 M’s, 3 A’s, and 2 T’s.
12!
= 19958400
2!3!2!
Circular Permutations
In general, the number of ways of arranging n objects around a circle is (n – 1)!
Question:
Solution:
At a graduation party, guests were seated in groups of 10 at circular
tables. How many permutations are there for each table?
(10 – 1)! = 9! = 362880
Permutation
the number of permutations of n objects taken r at a time
“n permute r”
n!
n Pr =
( n − r )!
Permutations with Repetitions =
n!
a!b!c!
Circular Permutations = (n – 1)!
4
Focus G
Determining the Number of Possible Combinations
In general, the number of combinations of n items taken r at a time is denoted by
the following formula:
P
n!
n Pr
= n r =
n Cr =
r! ( n − r )!r!
r Pr
Question:
Solution:
Question:
Solution:
Suppose 7 students from a class of 15 are to be chosen to go on a
school trip. In how many ways can they be selected?
15!
15!
=
= 6435
15 C 7 =
(15 − 7)!7! 8!7!
How many class committees of 4 students can be formed from these
15 students?
15 C 4 = 1365
Diagonals and Polygons
Question:
How many diagonals are there in a hexagon?
Solution:
Choose 2 points out of 6 to be joined, then don’t count adjacent pairs.
6 C 2 − 6 = 15 − 6 = 9
n( n − 3)
n C2 − n =
2
Combination
the number of combinations of n objects taken r at a time
“n choose r”
n!
n Cr =
( n − r )! r!
Diagonals in a Polygon
n C2 − n
5
5.3 Extra Practice
1.
How many permutations are there of all the letters in BANANA?
2.
Find the number of arrangements of the letters of the word TATTOO under
each condition:
a)
begins with a T
b)
begins with two T’s
c)
begins with three T’s
d)
begins with one T and the next letter is not a T
e)
begins with exactly two T’s
3.
A donut shop offers 9 varieties of donuts. Suppose a customer orders half a
dozen assorted donuts. How many different combinations are there?
4.
A volleyball team has 12 members.
a)
In how many ways can the coach choose the starting line-up of 6
players?
b)
In how many ways can the coach position the 6 starting players?
5.
From a deck of 52 cards, how many different 5-card hands can be formed in
each case?
a)
with exactly 3 aces
b)
with fewer than 3 aces
c)
with more than 3 aces
6.
a)
b)
7.
A customer chooses 5 or 6 tapes from a bin of 40. What is the expression
that gives the total number of possibilities?
8.
In how many orders can 7 students arrange themselves
a)
in a straight line?
b)
around a circular table?
9.
In how many different orders of colour can three white and three red roses
be arranged
a)
in a row?
b)
in a circular flower arrangement?
How many diagonals are there in a regular polygon with 13 sides?
There are 170 diagonals in a polygon. Determine how many sides are
on the polygon.
6
10.
The number of ways that a selection of 7 students can be chosen from a
class of 28 is the same as the number of ways that n students can be chosen
from the same class. What is the value of n?
11.
A group of 4 journalist is to be chosen to cover a murder trial. There are 5
male and 7 female journalists available. How many possible groups can be
formed:
a)
consisting of 2 men and 2 women?
b)
consisting of at least one woman?
12.
Collinear points are points which share the same straight line. How many
triangles are determined by 10 point, if no three of the points are collinear?
Answers
1. 60; 2. a) 30 b) 12 c) 3 d) 18 e) 9; 3. 84; 4. a) 924 b) 720; 5 a) 4512 b) 2594400 c) 48; 6. a) 65 b) 20; 7. 40C 5 + 40C 6; 8. a)
5040 b) 720; 9. a) 20 b) 10; 10. 21; 11. a) 210 b) 490; 12. 120
7