Section 6.4 Permutations and Combinations: Part 1

Section 6.4 Permutations and Combinations: Part 1

Permutations Permutations

1. How many ways can you arrange three people in a line?

2. Five people are waiting to take a picture. How many ways can you arrange three of them in a

line to take the picture?

Reminder of Factorials

n-Factorial (n!) For any natural number n,

n! = n(n� 1)(n� 2) · · · 3 · 2 · 1

0! = 1

Calculator Steps: Enter the number followed by MATH , scroll right to PRB and click 4 then

click ENTER .

Permutations of n Distinct Objects:

The number of permutations of n distinct objects taken r at a time is

P (n, r) =n!

(n� r)!

Calculator Steps: Enter the first number followed by MATH , scroll right to PRB and click 2

then enter the second number and click ENTER .

Note: If n = r, then P (n, n) = n!

3. Compute P (3, 3) and P (5, 3). How do these two permutations relate to the answers in examples

1 and 2?

*- I° I eI =

6w= 3 !

-

.I =

600€= 5 n P r 3

=

-

- - - -

-

- s n ←-

-

-- -

-

- -

-- -

3 n Pr 3,

= 3o÷ = 31.

-

- ⑥

5 n P r 3 =

,.

a 5¥ = 5.4.3.az#=6oy

Note: In this class, I tend to use the Multiplication Principle interchangeably with Permutations.

4. A company car that has a seating capacity of eight is to be used by eight employees who have

formed a car pool. If only three of these employees can drive, how many possible seating arrange-

ments are there for the group?

5. There are four families attending a concert together. Each family consists of 1 male and 2 females.

In how many ways can they be seated in a row of twelve seats if

(a) There are no restrictions?

(b) Each family is seated together?

(c) The members of each gender are seated together?

2 Fall 2019, Maya Johnson

* *- - -

- -

--

-

• ÷ I. -7. -6.-5--4=3 .-2.1=3.7 !=I5

( 3nRr1 ) . ( 7nPr7)=3z÷ . 7 ! =I53

= - -

12 ! -

-479,001,600J

- Grouping Problem !

4fam,

3 in each family .

'e is ! )"

.4 !

arrant4 o 3 o 2 o

I

×¥µwB =31

2 groups- Grouping Problem

4 males

8 females

±e⇒ gets-4-3-4= 4 ! -8 ! . 2 !

-2. I

= 1,935,360J

6. At a college library exhibition of faculty publications, two mathematics books, four social science

books, and three biology books will be displayed on a shelf. (Assume that none of the books are

alike.)

(a) In how many ways can the nine books be arranged on the shelf?

(b) In how many ways can the nine books be arranged on the shelf if books on the same subject

matter are placed together?

Permutations of n Objects, Not all Distinct:

Given a set of n objects in which n1 objects are alike and of one kind, n2 objects alike and of

another kind, . . . , and nm objects are alike and of yet another kind, so that

n1 + n2 + · · ·+ nm = n

then the number of permutations of these n objects taken n at a time is given by

n!

n1!n2! · · ·nm!


-

=- - -

-

9 !

=362-

Grouping Problem-

3 groups2 Math

4 Soc .

Sci .

3 Bio =3 ! . 2 ! .4 ! . 3 !

= 11728L

- -

- -

-

T a * *

7. Find the number of distinguishable arrangements of each of the following “words.”

(a) acccdbens

(b) baaaben

(c) A “word” that has 4A’s, 5B’s, and 1P

8. A toy chest contains three identical blue blocks, five identical yellow blocks, and 4 identical red

blocks. How many distinguishable arrangements of these blocks can be made?

9. Suppose that in the previous example, the blocks of the same color are numbered, so that the

blue blocks are numbered 1 through 3, the yellow blocks are numbered 1 through 5 and the red

blocks are numbered 1 through 4. Note that this means that blocks of the same color are no

longer identical.

(a) How many distinguishable arrangements of these blocks can be made?

(b) How many distinguishable arrangements of these blocks can be made if blocks of the same

color should stay together?


- .

# of objects = 9 # of arrangements :

1a,

3C,

Id,

Ib,

te,

In,

Is 931,2=60,4807

# of objects =7

2b,

3A,

Ie,

In2¥31

.

= 4200

- - - 10 !

# of objects = 10-4¥

- -1,26€

- - -

--

# blocks =3 t 5+4=12

3 B

1212227,720J54 3 ! . -5 ! . 4 !

4 R

- -

-

-

12 ! --479,001,60€

- Grouping Problem-

3 groups±± €⇒±

-4-3-2t=3 ! . 31.41 .

-5 !3 B

- --

s y 3 Z "

= 1031680J4 R

Section 6.4 Permutations and Combinations Part 2

Question:

Suppose we want to choose three people from a group of four people and we do not care about the

order in which we do this, that is, we will not be arranging the people we choose in any particular

order. How do we do this?

Answer:

Suppose we number the people from 1 through 4 and think of the set A = {1, 2, 3, 4}. To answer

this question we will count how many of size 3 there are of this set...

Combinations of n Distinct Objects:

The number of combinations of n distinct objects taken r at a time is given by

C(n, r) =n!

r!(n� r)!(where r n)

Calculator Steps: Enter the first number followed by MATH , scroll right to PRB and click 3

then enter the second number and click ENTER .

10. Compute C(4, 3) and C(10, 5).

Language If a problem uses the word “and” (\) then you need to multiply the results. If a

problem uses the word “or” ([) then you need to add the results.


- - -

-

-

.

Tubsets-

{ 112133,52 . 3,43 , { I, 3,43 ,

St , 2,43

4 subsets of size 3 to

choose 3 people

from4 people

- -

no -

. O -

-- -

-

-

-

4 n Cr-3=40to n Cr 5 =

252J

- - -

-- -

11. In how many ways can a subcommittee of six be chosen from a Senate committee of six Democrats

and five Republicans if

(a) All members are eligible?

(b) The subcommittee must consist of three Republicans and three Democrats?

12. From a shipment of 25 transistors, 6 of which are defective, a sample of 9 transistors is selected

at random.

(a) In how many di↵erent ways can the sample be selected?

(b) How many samples contain exactly 3 defective transistors?

(c) How many samples contain no defective transistors?


- - -

-

GD , 5 R → Total 11, Choose 6

11 n Cr 6 = 462J

✓multiply

-

I-

( 5 n C - 3) . ( Gnc - 31=2000

- Estate-6 defee .

19 not defee.

-

25mL - 9 =

2,042,9€#add

¥09

( 6 a C - 3)'s ( 19 nc - 6) a 5421640J

.

defee .not defeo .

-

( 6 wer O ) . ( 19 n C- 9)

= ( 19 n Cr 9) = -92,378J

(d) How many samples contain at least 5 defective transistors?

Complement Rule: Sometimes it is easier to ask how many ways there are of doing the

opposite (or complement) of what you want than it is to ask how many ways there are to do what

you want. So the complement rule is

# of Ways You Want = Total Ways - # of Ways You Don’t Want

13. A box contains 8 red marbles, 8 green marbles, and 10 black marbles. A sample of 12 marbles is

to be picked from the box.

(a) How many samples contain at least 1 red marble?

(b) How many samples contain exactly 4 red marbles and exactly 3 black marbles?


5 or 6 defect .

-

5 defee .

6 defee .

( Gnc - 5) . ( 19nA 4) t ( Gnc - 6) . ( l9nCr3 )

724,225J-

-

- - -

total- 8k

86 choose

I Roc 2 Ro - . -- .8R LOB 12

No red → Don 't want

Total -

u don't wa -

two reds

( 26 nc - 12 ) - ( 18 NER12 ) = 9,6391136J

4th n 3 B A SG

- . -

Csn 'T4) . lioness . 1812-5 )

= 470,400J

(c) How many samples contain exactly 7 red marbles or exactly 6 green marbles?

(d) How many samples contain exactly 5 green marbles or exactly 3 black marbles?


U

7- RUGG = 2K +6 I

#> r

← %%e ? iz=Lsent 7) . ( Isner -5 )26 ↳#

+ GG

f 8nCr 6) . ( 18mL - 6)

2588,336J-

:-

5GU3B=5Gt3B - 5Gn3B A Hot

56

I ( 8h45 ) of 18h47 ) t

3B

. ( lower 3) ( 16h49 ) -

( 8nG E). ( Nnc - Z) .(8nCr4 )

= 2,684,544J

When To Use Permutations/Multiplication Principle or Combinations?

We use permutations/multiplication principle whenever order matters and we use combinations

whenever order does not matter.

Keywords that suggest we use permutations/multiplication principle: distinguishable ar-

rangments, ordered list of names, distinct arrangments, arrange in a line, seated in a line or

seated in any arrangment

Keywords that suggest we use combinations: choose a smaller group from a larger group,

select a committee or subcommittee, select a number of items, how many samples contain a

number of items or people

14. In how many ways can the names of three Republican and four Democratic candidates for political

o�ce be listed on a ballot?

(a) Should we use combinations or should we use permutations/multiplication principle?

(b) How many ways can this be done?

15. A bag contains 5 red marbles and 5 blue marbles. How many ways can you select 6 marbles from

the bag?

(a) Should we use combinations or should we use permutations/multiplication principle?

(b) How many ways can this be done?

Note: Sometimes a problem requires using both permutations/multiplication principle and com-

binations.


-

- - -

-

- - -

- - - -

-

I-

3 t 4 = 7

Malt . Pr .ci pie

7-IEe-€ = 71.

= 5,040J- - -

Combinations

Total 10

loner 6 = choose 6

16. Suppose we have 25 people on a committee. How many subcommittees contain one president,

one vice president and six cabinet members?

17. Twenty runners are competing in a half-marathon. How many ways can we award one 1st place

prize, one 2nd place prize, one 3rd place prize, and four 4th place prizes?


- - -

--

pie.

.

ftp.#E6I--60,568,200J

- -

- - -

1¥& •Ltd = 16,2792000

Documents

Section 6.4 Permutations and Combinations: Part 1