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?

4

• 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

• Determine the of arrangements of the following
• FOUR
• NINE
• RADII
• GAMMAS
• COMBINATION

76
Solution 3.12

• Determine the of arrangements of the following
• FOUR
• All the letters are unique, so 4! 24.

77
Solution 3.12

• Determine the of arrangements of the following
• (b) NINE
• 1st count the arrangements if the two Ns are
• unique, which is 4!. Then since the Ns are not
• unique, divide by 2!. The answer is 4! / 2! 12.

78
Solution 3.12

• Determine the of arrangements of the following
• (c) RADII
• 1st account for the Is as if they are unique,
  5!. Then
• since the Is are not unique, divide by 2! so the
  
• answer is 5! / 2! 60.

79
Solution 3.12

• Determine the of arrangements of the following
• (d) GAMMAS
• There are 2 As, 2 Ms, and six total letters, so
• 6! / (2! x 2!) 180.

80
Solution 3.12

• Determine the of arrangements of the following
• (e) COMBINATION
• Eleven total letters, two Os, two Is, and two
  Ns.
• 11! / (2! x 2! x 2!) 4,989,600.

81
Review 3.13

• I have 3 identical math books, 3 identical
  English
• books, and 2 identical French books. In how many
• ways can I arrange them on the shelf is all I
  care
• about is the order of the subjects (in other
  words,
• all 3 math books are considered the same)?

82
Solution 3.13

• I have 3 identical math books, 3 identical
  English
• books, and 2 identical French books. In how many
• ways can I arrange them on the shelf is all I
  care
• about is the order of the subjects (in other
  words,
• all 3 math books are considered the same)?
• There are 8! ways to arrange the books if they
  are
• unique, but we must divide out the permutations
• of identical books, so
• 8! / (3! x 3! x 2!)
  560.

83
Review 3.14

• In how many ways can the digits 45, 520 be
• arranged to form a 5-digit number?

84
Solution 3.14

• In how many ways can the digits 45, 520 be
• arranged to form a 5-digit number?
• 1st place the 0, which has only 4 options (0
  cannot
• be the 1st digit). Then there are 4 remaining
  places
• to put the last 4 digits, two of which are not
• unique (5s), so there are 4! / 2! options for
• arranging the other 4 digits. The answer is
• (4 x 4!) / 2! 48.

85
Review 3.18

• There are 6 married couples at a party. At the
  start
• of the party, every person shakes hands once with
• every other person except with his or her spouse.
• How many handshakes are there?

86
Solution 3.18

• There are 6 married couples at a party. At the
  start
• of the party, every person shakes hands once with
• every other person except with his or her spouse.
• How many handshakes are there?
• All 12 people shake hands with 10 other people
• (everyone except themselves and their spouses).
  In
• muliplying 12 x 10, each handshake is counted
• twice, so we divide by two. (12 x 10) / 2 60.

87
Review 3.19

• Seven points are marked on the circumference of
• a circle. How many different chords can be drawn
• by connecting two of these seven points? (Source
• MATHCOUNTS).

88
Solution 3.19

• Seven points are marked on the circumference of
• a circle. How many different chords can be drawn
• by connecting two of these seven points? (Source
• MATHCOUNTS).
• Choose two out of seven points (without regard to
• order) in (7 x 6) / 2 21 ways, so there are 21
• chords.

89
Review 3.20

• How many pairs of vertical angles are formed by
• five distinct lines that have a common point of
• intersection? (Source MATHCOUNTS)

90
Solution 3.20

• How many pairs of vertical angles are formed by
• five distinct lines that have a common point of
• intersection? (Source MATHCOUNTS)
• Each pair of lines will give two pairs of
  vertical
• angles. There are (5 x 4) / 2 10 ways to choose
  a
• pair of lines, therefore there are 2 x 10 20
  pairs
• of vertical angles.