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Solution 3.4.4 In how many ways can we seat 6 people around a table if Fred and Gwen insist on sitting opposite each other? – PowerPoint PPT presentation

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Title: AoPS:

1
AoPS

Introduction to Probability and Counting

2
Chapter 3

Correcting
for
Overcounting

3
Permutations with Repeated Elements

Lets start with a problem we should
(hopefully) already know how to do.
Problem 3.1 How many possible
distinct arrangements are there of the letters in
the word DOG?

Problem 3.1
How many possible distinct arrangements are
there of the letters in the word DOG?
We could list them or we should know by now
that there are 3 ways to pick the 1st, 2 ways to
pick the 2nd, and 1 way to pick the last letter,
for a total of 3! 6 ways.
This is just a basic permutation problem like
in Chapter 1.

5
Problem 3.2

How many possible distinct arrangements are
there of the letters in the word BALL?

6
Solution 3.2

How many possible distinct arrangements are
there of the letters in the word BALL?
The Bogus Solution 4! possibilities.
Thats because BAL1L2 and BAL2L1 are the same
once the subscripts are removed, so we have
over-counted and need to correct this.

7
Solution 3.2

How many possible distinct arrangements are
there of the letters in the word BALL?
The Bogus Solution is 4! Possibilities.
Thats because BAL1L2 and BAL2L1 are the same
once the subscripts are removed, so we have
over-counted and need to correct this. So we have
to divide the of arrangements by 2! because the
Ls can be arranged 2 different ways 4! / 2!
12

8
Problem 3.3

How many distinct arrangements are there of
TATTER?

9
Solution 3.3

TATTER pretend that all the Ts are different
(T1, T2, and T3). Then are are 6! possibilities.
The 3 Ts can be arranged in 3! different ways,
which leads to overcounting.
The solution 6! / 3! 120.
This is called strategic overcounting.

10
Problem 3.4

One more twist
How many distinct arrangements are there of
PAPA?

11
Solution 3.4

One more twist
How many distinct arrangements are there of
PAPA?
Simple because the Ps repeat twice and the As
repeat twice, the answer is
4! / (2! X 2!) 6 ways

12
Exercise 3.1

Compute the of distinct arrangements of the
letters in the word EDGE.

13
Solution 3.1

Compute the of distinct arrangements of the
letters in the word EDGE.
4! / 2! 12 ways

14
Exercise 3.2

For each of the following words, determine the
of ways to arrange the letters of the word.
WAR
THAT
CEASE
ALABAMA
MISSISSIPPI

15
Solution 3.2

For each of the following words, determine the
of ways to arrange the letters of the word.
WAR 3! 6
THAT 4! / 2! 12
CEASE 5! / 2! 60
ALABAMA 7! / 4! 210
MISSISSIPPI 11! / (4! X 4! X 2!) 34,650

16
Problem 3.3

I have 5 books, two of which are identical
copies of the same math book (and all of the rest
of the books are different.) In how many ways can
I arrange them on the shelf?

17
Solution 3.3

I have 5 books, two of which are identical
copies of the same math book (and all of the rest
of the books are different.) In how many ways can
I arrange them on the shelf?
5! / 2! 60

18
Problem 3.4

There are 8 pens along a wall in the pound. The
pound has to allocate 4 pens to dogs, 3 to cats,
and one to roosters. In how many ways can the
pound make the allocation?

19
Solution 3.4

There are 8 pens along a wall in the pound. The
pound has to allocate 4 pens to dogs, 3 to cats,
and one to roosters. In how many ways can the
pound make the allocation?
8! / (4! X 3!) 280

20
Counting Pairs of Items

A round-robin tennis tournament consists of each
player playing every other player exactly once.
How many matches will be held during the an 8-
person round-robin tournament?

21
Counting Pairs of Items

A round-robin tennis tournament consists of each
player playing every other player exactly once.
How many matches will be held during the an 8-
person round-robin tournament?
Bogus Solution Each of the 8 players plays 7
games, or 8 x 7 56 total games played.

22
Counting Pairs of Items

A round-robin tennis tournament consists of each
player playing every other player exactly once.
How many matches will be held during the an 8-
person round-robin tournament?
Alice plays Bob and Bob plays Alice, but its the
same game, so 8 x 7 28.
2

23
Counting Pairs of Items

Another way to look at the problem Alice plays
7 matches. Bob also plays 7 matches, but the one
with Alice has already been counted, leaving
Bob with 6 more to play. The next player, Carol,
has already played A B, so Carol has 5 more
matches to play, and so on down the line, to the
last player.

24
Counting Pairs of Items

Another way to look at the problem Alice plays
7 matches. Bob also plays 7 matches, but the one
with Alice has already been counted, leaving
Bob with 6 more to play. The next player, Carol,
has already played A B, so Carol has 5 more
matches to play, and so on down the line, to the
last player. That means we have a total of
7 6 5 4 3 2 1 28 games total played

25
Counting Pairs of Items

That means we have a total of
7 6 5 4 3 2 1 28 games total
played.
This is a classic example of counting pairs of
objects.

26
Problem 3.5

Compute the sum 1 2 3 4 of the 1st 4
positive integers.
Compute the sum of 1 2 3 4 5 of the 1st 5
positive integers.
Compute the sum of 1 2 3 9 10 of the
1st 10 positive integers.
Find a formula for the sum of the 1st n positive
integers.
Compute the sum of 1 2 3 100 of the 1st
100 positive integers.

27
Solution 3.5

Compute the sum 1 2 3 4 of the 1st 4
positive integers. 10
Compute the sum of 1 2 3 4 5 of the 1st 5
positive integers. 15
Compute the sum of 1 2 3 9 10 of the
1st 10 positive integers. 55
Find a formula for the sum of the 1st n positive
integers. n (n 1) / 2
Compute the sum of 1 2 3 100 of the 1st
100 positive integers. 5050

28
Problem 3.6

A round-robin tennis tournament consists of each
player playing every other player exactly once.
How many matches will be held during a n-person
round-robin tennis tournament, where n gt 2 is a
positive integer?

29
Solution 3.6

A round-robin tennis tournament consists of each
player playing every other player exactly once.
How many matches will be held during a n-person
round-robin tennis tournament, where n gt 2 is a
positive integer?
Each of the n players must play every other
player, so each player must play n 1 matches
which
gives the preliminary count n (n 1) matches.

30
Solution 3.6

Each of the n players must play every other
player, so each player must play n 1 matches
which
gives the preliminary count n (n 1) matches.
But
this counts each match twice, so divide by 2 to
get
n (n 1)
2

31
Problem 3.7

A convex polygon is a polygon in which every
interior angles is less than 180. A diagonal of
a
convex polygon is a line segment which connects
2 non-adjacent vertices. Find a formula for the
of diagonals of a convex polygon with n sides,
where n is any positive integer greater than 2.

32
Solution 3.7

A polygon with n sides has n vertices. A diagonal
corresponds to a pair of vertices. By similar
reasoning to Prob. 3.6, there are n (n 1)
pairs
2
of vertices.

33
Solution 3.7

A polygon with n sides has n vertices. A diagonal
corresponds to a pair of vertices. By similar
reasoning to Prob. 3.6, there are n (n 1)
pairs
2
of vertices. However n of these pairs correspond
to edges of the polygons rather than diagonals,
so
subtract these from the count the of diagonals
n (n 1) - n
2

34
Solution 3.7

Simplifying this expression n (n 1) - n
2
leads to the following
n (n 3)
2
If youre interested in the math behind this
transformation, I can show it to you all you
have to do
is ask!

35
Exercise 3.3.1

A club has 15 members and needs to choose 2
members to be co-presidents. In how many ways
can the club choose its co-presidents?

36
Solution 3.3.1

A club has 15 members and needs to choose 2
members to be co-presidents. In how many ways
can the club choose its co-presidents?
If the co-president positions are unique, there
are
15 choices for the 1st and 14 choices for the
2nd.
However, since the positions are identical, we
must divide by 2! (the of arrangements of co-
presidents) (15 x 14) / 2! 105 ways.

37
Exercise 3.3.2

I have twenty balls numbered 1 through 20 in a
bin. In how many ways can I select 2 balls is the
order in which I draw them doesnt matter?

38
Solution 3.3.2

I have twenty balls numbered 1 through 20 in a
bin. In how many ways can I select 2 balls is the
order in which I draw them doesnt matter?
This is like the previous problem, so we get
(20 x 19) / 2! 190

39
Exercise 3.3.3

A sports conference has 14 teams in two divisions
of 7. How many games are in a complete season
for the conference if each team must play every
other team in its own division twice and every
team in the other division once?

40
Solution 3.3.3

A sports conference has 14 teams in two divisions
of 7. How many games are in a complete season
for the conference if each team must play every
other team in its own division twice and every
team in the other division once?
Each team plays 6 other teams in its division
twice
and the other 7 teams once, for a total of 6 x 2
7
19 games for each team.

41
Solution 3.3.3

Each team plays 6 other teams in its division
twice
and the other 7 teams once, for a total of 6 x 2
7
19 games for each team.
There are 14 teams total, which gives a
preliminary
count of 19 x 14 266 games, but we must divide
by 2 because we counted each game twice. The
answer is (19 x 14) / 2 133 games.

42
Exercise 3.3.4

Find a formula for the sum of the 1st n even
integers 2 4 6 2n.
Find a formula for the sum of the 1st n odd
integers 1 3 5 (2n 1).

43
Solution 3.3.4

Find a formula for the sum of the 1st n even
integers 2 4 6 2n.
Let S 2 4 6 2n. ( S had n terms.)
Another expression for S is 2n 6 4 2, so
adding these together we obtain the equation
2S (2n 2) (2n 2) (2n 2) .

44
Solution 3.3.4

Find a formula for the sum of the 1st n even
integers 2 4 6 2n.
Let S 2 4 6 2n. ( S had n terms.)
Another expression for S is 2n 6 4 2, so
adding these together we obtain the equation
2S (2n 2) (2n 2) (2n 2) . This
sum
has n terms, so 2S n (2n 2) 2n (n 1),
and
S n (n 1)

45
Solution 3.3.4

Let S 2 4 6 2n. ( S had n terms.)
Another expression for S is 2n 6 4 2, so
adding these together we obtain the equation
2S (2n 2) (2n 2) (2n 2) . This
sum
has n terms, so 2S n (2n 2) 2n (n 1),
and
S n (n 1)
OR, S 2(1 2 3 n) 2(n(n 1))
2
n (n 1).

46
Solution 3.3.4

(b) Find a formula for the sum of the 1st n odd
integers 1 3 5 (2n 1).
Let S 1 3 5 (2n 1). (S has n
terms.)
Another expression for S (2n 1) 3 1,
so adding these together we get the equation
2S 2n 2n 2n.

47
Solution 3.3.4

(b) Find a formula for the sum of the 1st n odd
integers 1 3 5 (2n 1).
Let S 1 3 5 (2n 1). (S has n
terms.)
Another expression for S (2n 1) 3 1,
so adding these together we get the equation
2S 2n 2n 2n.
This sum has n terms, so 2S n (2n) 2n2 so
S n2.

48
Exercise 3.3.5

How many interior diagonals does an icosahedron
have? ( An icosahedron is a 3-dimentional figure
with 20 triangular faces and 12 vertices, with 5
faces meeting at each vertex. An interior
diagonal
is a segment connecting two vertices which do not
lie on a common face.)

49
Solution 3.3.5

How many interior diagonals does an icosahedron
have?
There are 12 vertices in the icosahedron, so
there
are potentially 11 other vertices to which we
could
extend a diagonal. However 5 of these 11 points
are connected to the original point by an edge,
so
they are not connected by internal diagonals.

50
Solution 3.3.5

There are 12 vertices in the icosahedron, so
there
are potentially 11 other vertices to which we
could
extend a diagonal. However 5 of these 11 points
are connected to the original point by an edge,
so
they are not connected by internal diagonals.
So each vertex is connected to 6 other points by
interior diagonals. This gives the preliminary
count of 12 x 6 72 interior diagonals. However,
they were counted twice, so divide by 2 36
diag.

51
Counting with Symmetries

Problem 3.8 In how many ways can 6 people
be seated at a round table? Two seating arrange-
ments are considered the same if, for each
person,
the person to his left or right is the same in
both
arrangements. In other words, the 2 arrangements
shown are the same.

C
A
F
B
B
D
C
A
E
E
D
F
52
Counting with Symmetries

Solution 3.8 If 6 people were sitting in a
row,
there would be 6! arrangements. But this is a
case
of overcounting. The reason for this is the
problem
has rotational symmetry therefore, we must
divide
the overcount by 6, so the answer is
6! / 6 5! 120

C
A
F
B
B
D
C
A
E
E
D
F
53
Counting with Symmetries

Solution 3.8 There is another way to solve
the
problem, using a constructive counting approach.
First place person A. Since all rotations of the
same seating are considered identical, person A
is
essentially fixed the rotation. Place the rest
of
the people in the usual way.

C
A
F
B
B
D
C
A
E
E
D
F
54
Counting with Symmetries

Solution 3.8 There are 5 choices for where
to
place person B, 4 remaining choices for where to
place person C, etc., for a total of
5 x 4 x 3 x 2 x 1 5! 120 possible seatings.

C
A
F
B
B
D
C
A
E
E
D
F
55
Exercise 3.4.1

In how many ways can 8 people be seated
around a round table?

56
Solution 3.4.1

In how many ways can 8 people be seated
around a round table?
There are 8! ways to place the people around the
table, but this counts each valid arrangement 8
times (once for each rotation of the same
arrange-
ment). The answer is 8! / 8 7! 5040.

57
Solution 3.4.2

In how many ways can 5 keys be placed on a key-
chain?
There are 5! ways to place the keys on the key-
chain, but we must divide by 5 for rotational
symmetry (5 rotations for each arrangement) and
by 2 for reflectional symmetry (flip the keychain
to get the same arrangement).
5! / (5 x 2)
12.

58
Exercise 3.4.3

A Senate committee has 5 Democrats and 5
Republicans. In how many ways can they sit
around a circular table
without restrictions?
if all the members of each party all sit next to
each other?
if each member sits next to members of the other
party?

59
Solution 3.4.3

A Senate committee has 5 Democrats and 5
Republicans. In how many ways can they sit
around a circular table
without restrictions?
There are 10 people to place, so place them in
10!
ways, but this counts each valid arrangement 10
times (once for each rotation of the same
arrange-
ment). So the of ways to seat them is
10! / 10 9! 362,880.

60
Solution 3.4.3

A Senate committee has 5 Democrats and 5
Republicans. In how many ways can they sit
around a circular table
(b) if all the members of each party all sit next
to each other?
Choose any 5 seats in which to seat the Democrats
it doesnt matter which 5 consecutive seats we
choose. Then there are 5! ways to place the Ds
and
5! ways to place the Rs in their seats for the
total
5! x 5! 14, 400.

61
Solution 3.4.3

(c) if each member sits next to members of the
other party?
The only way the Senators can be seated is if the
seats alternate by party.

D
R
R
D
D
R
R
D
D
R
62
Solution 3.4.3

Fix the rotation by placing the youngest Democrat
in the top seat, so that we have removed the
over-
counting of rotations of the same arrangement.

D
R
R
D
D
R
R
D
D
R
63
Solution 3.4.3

Now there are 4! ways to place the Democrats left
in the other Democrat seats, and 5! ways to place
the
Republicans in the Republican seats, for a total
of
5! x 4! 2,880 arrangements.

D
R
R
D
D
R
R
D
D
R
64
Problem 3.4.4

In how many ways can we seat 6 people around
a table if Fred and Gwen insist on sitting
opposite
each other?

65
Solution 3.4.4

In how many ways can we seat 6 people around
a table if Fred and Gwen insist on sitting
opposite
each other?
There are 6 choices of seats for Fred to sit in.
Once Fred is seated, then Gwen must sit opposite
him. This leaves 4 people to place in the four
remaining seats, which can be done 4! ways.

66
Solution 3.4.4

There are 6 choices of seats for Fred to sit in.
Once Fred is seated, then Gwen must sit opposite
him. This leaves 4 people to place in the four
remaining seats, which can be done 4! ways.
However, we must divide by 6 to account for the 6
rotations of the table. So the of arrangements
is
(6 x 1 x 4!) / 6 4! 24.

67
Problem 3.4.5

In how many ways can we seat 8 people around a
table if Alice and Bob wont sit next to each
other?

68
Solution 3.4.5

In how many ways can we seat 8 people around a
table if Alice and Bob wont sit next to each
other?
There are 8 choices for seats for Alice. Once
Alice
is seated, there are 5 seats left for Bob, since
he
wont sit in either seat immediately next to
Alice.
This leaves 6 people to place in the remaining 6
seats, which can be done 6! ways.

69
Solution 3.4.5

There are 8 choices for seats for Alice. Once
Alice
is seated, there are 5 seats left for Bob, since
he
wont sit in either seat immediately next to
Alice.
This leaves 6 people to place in the remaining 6
seats, which can be done 6! ways.
However, we must divide by 8 to account for the 8
rotations of the table. So the of arrangements
is
(8 x 5 x 6!) / 8 5 x 6! 3600.

70
Review 3.11

How many arrangements are there of ste1e2e3
(consider e1,e2,e3 to be different letters)?
(b) List the arrangements of ste1e2e3 which
have
st as the 1st two letters. How many are there?
(c) List the arrangements of steee (The es are
all
the same this time.) How many are there?
(d) Let p be the answer to (a), q be the answer
to (b) and r be the answer to (c). Is p/q r? If
so, why must it be so? If not, why not?

71
Solution 3.11

How many arrangements are there of ste1e2e3
(consider e1,e2,e3 to be different letters)?
This is the of ways to arrange 5 unique
objects,
which is 5! 120.

72
Solution 3.11

(b) List the arrangements of ste1e2e3 which
have
st as the 1st two letters. How many are there?
ste1e2e3, ste1e3e2, ste2e1e3, ste2e3e1, ste3e1e2,
and
ste3e2e1, giving 6 arrangements.

73
Solution 3.11

(c) List the arrangements of steee (The es are
all
the same this time.) How many are there?
steee, setee, seete, seeet, estee, esete, eseet,
eeste,
eeset, eeest, tseee, tesee, teese, teees, etsee,
etese,
etees, eetse, eetes, eeets, giving 20
arrangements.

74
Solution 3.11

(d) Let p be the answer to (a), q be the answer
to (b) and r be the answer to (c). Is p/q r? If
so, why must it be so? If not, why not?
Yes, we see that 120/6 20. The q counts the
number of ways that each arrangement from part
(c) is overcounted in part (a), so we must divide
p by q to get the number of arrangements r in
part
(c).

75
Review 3.12