AoPS:
Introduction to
Probability and
Counting
Chapter 4
Committees and
Combinations
Committee Forming
Problem 4.1
(a)In how many ways can a President and a Vice
President be chosen from a group of 4 people
(assuming that the President and the Vice-President cannot be the same person)?
Committee Forming
Problem 4.1
This is a permutation problem like those in Chp1.
There are 4 choices for Pres. and 3 choices for VP,
so there are 4 x 3 = 12 choices.
Committee Forming
Problem 4.1
(b) In how many ways can a 2-person committee be
chosen from a group of 4 people (where the order in
which we choose the 2 people doesn’t matter)?
Committee Forming
Problem 4.1
There are 4 ways to choose person A & 3 ways to
choose person B, but then we’ve overcounted again
since choosing person A and then person B will give
the same committee as choosing person B and then
person A. Each committee is counted twice in the
original 4 x 3 count, so divide by 2 to correct for
overcount: (4 x 3) /2 = 6 ways to choose the 2 person
committee.
Committee Forming
Problem 4.1
Not convinced? Label the people, A, B, C, & D.
Then list all the possible ways to select a President
& Vice-President.
Committee Forming
Pres. VP
A
B
B
A
A
C
C
A
A
D
D
A
Problem 4.1
Committee
Pres. VP Committee
AB
B
C
BC
C
B
AC
B
D
BD
D
B
AD
C
D
CD
D
C
Important: Understand the difference between forming
committees (order doesn’t matter) & picking distinct
officers (order of selection matters).
Pres. VP Committee
A
B
AB
B
A
A
C
AC
C
A
A
D
AD
D
A
Pres. VP Committee
B
C
BC
C
B
B
D
BD
D
B
C
D
CD
D
C
Problem 4.2
In how many ways can 3 people be chosen from
a group of 8 people to form a committee?
Problem 4.2
There are 8 ways to choose the 1st, 7 ways to choose
the 2nd, and 6 ways to choose the 3rd. But since we
are choosing a committee, not officers, the order
does
not matter. This is the key to any committee-forming
problems.
Problem 4.2
Suppose the committee is made up of A, B, & C. It
can be chosen in any of the following orders:
ABC, ACB, BAC, BCA, CAB, CBA
so each possible committee corresponds to 3! = 6
possible orderings of officers. Since each committee
is overcounted by 3!, the # of committees that can be
chosen from 8 people is
(8 x 7 x 6) / 3! = 56.
Combinations
Special notations are used for choosing committees.
Denote the # of ways we can choose an r-person
committee from a total of n people as C(n, r) or
n
C
or
n
r
r
Combinations
Special notations are used for choosing committees.
Denote the # of ways we can choose an r-person
committee from a total of n people as C(n, r) or
n
C
or
n
r
r
This is read as “n choose r” for the # of ways to
choose an r-person committee from a total of n people.
Combinations
Looking back at Problems 4.1 and 4.2
4
2
=6
8
3
= 56
Problem 4.3
In my state’s lottery, 48 balls are numbered 1-48,
and 6 are chosen. How many different sets of
winnings numbers are there? (In this lottery, the
order in which the numbers are chosen does not
matter.)
Problem 4.3
This is the same as forming a committee – instead
of people, we have balls.
48 x 47 x 46 x 45 x 44 x 43 = 12,271,512
6!
Problem 4.3
This is the same as forming a committee – instead
of people, we have balls.
48 x 47 x 46 x 45 x 44 x 43 = 12,271,512
6!
This number is also 48
6
This means your chance of winning the lottery is
1 in 12 million.
Exercises
4.2.1
(a) In how many ways can I choose 4
different officers from a club of 9 people?
(b) In how many ways can I choose a 4person committee from a club of 9 people?
Exercises
4.2.1
(a) In how many ways can I choose 4
different officers from a club of 9 people?
3024
(b) In how many ways can I choose a 4person committee from a club of 9 people?
126
Exercises
4.2.2 My club has 25 members. In how many
ways can I choose members to form a 4person executive committee?
Exercises
4.2.2 My club has 25 members. In how many
ways can I choose members to form a 4person executive committee?
12, 650
Exercises
4.2.3 Our water polo team has 15 members. I want
to choose a starting line-up consisting of 7
players, one of whom will be the goalie (the
other six positions are interchangeable). In how
many ways can I choose my starting line-up?
Exercises
4.2.3 Our water polo team has 15 members. I want
to choose a starting line-up consisting of 7
players, one of whom will be the goalie (the
other six positions are interchangeable). In how
many ways can I choose my starting line-up?
45,045
Exercises
4.2.4 Consider a regular octagon. How many
triangles can be formed whose vertices are the
vertices of the octagon?
Exercises
4.2.4 Consider a regular octagon. How many
triangles can be formed whose vertices are the
vertices of the octagon?
No 3 vertices are collinear so any combination of 3
vertices will from a triangle, so there are
8 x 7 x 6 = 56
3!
Exercises
4.2.5 The Senate has 100 members, consisting of
55 Republicans and 45 Democrats. In how many
ways can I choose a 5-person committee
consisting of 3 Republicans and 2 Democrats?
Exercises
4.2.5 The Senate has 100 members, consisting of
55 Republicans and 45 Democrats. In how many
ways can I choose a 5-person committee
consisting of 3 Republicans and 2 Democrats?
There are 55 x 54 x 53 = 26,235 ways to choose
R’s
3!
= 990
and 45 x 44
ways to choose D’s.
2!
So there are 26,235 x 990 = 25,972,650 ways to
Exercises
4.2.6 My state’s lottery has 30 white balls
numbered 1 – 30, and 20 red balls numbered 1-20.
In each lottery drawing, 3 of the white balls and 2
of the red balls are drawn. To win you must match
all 3 white balls and both red balls, without regard
to the order in which they were drawn. How many
possible different combinations may be drawn?
Exercises
4.2.6 My state’s lottery has 30 white balls
numbered 1 – 30, and 20 red balls numbered 1-20.
In each lottery drawing, 3 of the white balls and 2
of the red balls are drawn. To win you must match
all 3 white balls and both red balls, without regard
to the order in which they were drawn. How many
possible different combinations may be drawn?
Ways to choose white balls: 30 x 29 x 28 = 4060
3!
Ways to choose red balls: 20 x 19 = 190
2!
Exercises
4.2.6 My state’s lottery has 30 white balls
numbered 1 – 30, and 20 red balls numbered 1-20.
In each lottery drawing, 3 of the white balls and 2
of the red balls are drawn. To win you must match
all 3 white balls and both red balls, without regard
to the order in which they were drawn. How many
possible different combinations may be drawn?
So total # of outcomes for both red and white balls is
4060 x 190 = 771,400.
How To Compute Combinations
Problem 4.4: Consider a club which has n people.
(a)Determine a formula for the number of ways to
choose r different officers from n people.
(b) Determine the number of ways that any given
r
people can be assigned to be officers.
(c) Using your answers to parts (a) & (b),
determine a formula for the number of ways to
form an r-person committee from a total of n
How To Compute Combinations
Problem 4.4: Consider a club which has n people.
Start by counting the # of ways to choose r people
if order matters. There are n choices for the 1st
person, n – 1 choices for the 2nd person, n – 2
choices for the 3rd, and so on, up to n – r + 1
choices (See p.20, Prob 1.15 of Intro to Counting
& Probability.)
How To Compute Combinations
Problem 4.4: Consider a club which has n people.
So there are
n x (n – 1) x (n – 2) x … x (n – r + 1)
ways to choose r people from a total of n people
if order matters. This is the quantity that we denote
by P(n, r).
How To Compute Combinations
Problem 4.4: Consider a club which has n people.
But we know that there are r! ways to order r
people. Therefore, each unordered committee of r
people will correspond to r! ordered choices of r
people. So we need to divide our count by r! to
correct for overcounting.
How To Compute Combinations
n
r
n x (n – 1) x (n – 2) x … x (n – r + 1)
=
r!
n!
=
r! (n – r)!
Problem 4.5
Compute
11
4
Problem 4.5
Compute
11
4
= 11! = 11 x 10 x 9 x 8
4!7!
4x3x2x1
= 330
SHORTCUT: Put the 1st r terms of n! in the
numerator and r! in the denominator.
Problem 4.6
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 > 2?
Problem 4.6
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 > 2?
Use combinations. Think of each match in the
tournament as a 2-person committee, where the 2
players in the match are a “committee.”
Problem 4.6
Use combinations. Think of each match in the
tournament as a 2-person committee, where the 2
players in the match are a “committee.”
So the # of matches in the tournament equals the #
of ways to choose a 2-person committee out of n
people, which is n
2
Exercises
4.3.1 Compute the following combinations:
5
8
10
9
9
3
2
4
8
1
Exercises
4.3.1 Compute the following combinations:
5
8
9
10
28 10 210 9
9
3
2
4
8
1
9
Problem 4.3.2
What is
n
0
for any positive integer n?
Problem 4.3.2
What is
n
for any positive integer n?
0
0! = 1 by definition, so the combination equals
n!
=1
0! n!
Problem 4.3.2
What is
n
1
for any positive integer n?
Problem 4.3.2
What is
n
for any positive integer n?
1
n!
=n
1!(n -1)!
Problem 4.3.2
What is
n
n
for any positive integer n?
Problem 4.3.2
What is
n
for any positive integer n?
n
n!
=
n!(n -n)!
n!
n!0!
n! = 1
=
n!
Our First Combinatorial Identity
Problem 4.7: Compute
6
(a) 6
2 and
4
(b)
8
3
8
and
5
Our First Combinatorial Identity
Problem 4.7: Compute
6
(a) 6
2 and
4
6
2
= 6 x 5 = 15
2x1
6
4
= 6 x 5 x 4 x 3 = 15
4x3x2x1
Our First Combinatorial Identity
Problem 4.7: Compute
(a) 8 = 8 x 7 x 6 = 56
3
3x2x1
(b) 8 = 8 x 7 x 6 x 5 x 4
5
5x4x3x2x1
= 56
Problem 4.10
Compute
(a) 9
0
Problem 4.10
Compute
9
0
= 9! = 1
9!0!
Review Problems
4.12 Joe has one book each for algebra, geometry,
history, psychology, Spanish, English, and physics
in his locker. If he chooses three books from his
locker, how many different sets of three books
could he choose? (Source: MATHCOUNTS)
4.12 Joe has one book each for algebra, geometry,
history, psychology, Spanish, English, and physics
in his locker. If he chooses three books from his
locker, how many different sets of three books
could he choose? (Source: MATHCOUNTS)
Since we don’t care about the order that the books
are chosen, we have 7C3 = 35 possibilities.
Problem 4.13
I won a trip for four to the Super Bowl. I can bring
three of my friends. I have 8 friends. In how many
ways can I form my Super Bowl party?
Problem 4.13
I won a trip for four to the Super Bowl. I can bring
three of my friends. I have 8 friends. In how many
ways can I form my Super Bowl party?
Order does not matter, so it is a combination.
Choosing 3 out of 8 is 8C3 = 56.
Problem 4.14
A lock has 5 buttons numbered 1-5. The lock is
opened by pushing two buttons simultaneously and
then pushing one button alone. How many
combinations are possible?
(Source: MATHCOUNTS)
A lock has 5 buttons numbered 1-5. The lock is
opened by pushing two buttons simultaneously and
then pushing one button alone. How many
combinations are possible?
(Source: MATHCOUNTS)
There are 5C2 = 10 ways to choose the 1st two
buttons, and then 5 choices for the last button. So
there are 10 x 5 = 50 possible combinations.
Problem 4.15
I want to read 4 books over the next month. My
bookshelf has 12 books. In how many ways can I
choose which books to read over the next month,
without regard to the order that I read them?
Problem 4.15
Order does not matter, so it is a combination .
Choosing 4 from 12 is 495.
Problem 4.16
How many triangles can be formed using the
vertices of a regular dodecagon (12-sided polygon)?
Problem 4.16
How many triangles can be formed using the
vertices of a regular dodecagon (12-sided polygon)?
No three vertices are collinear, so any combination
of 3 vertices will make a triangle. Choosing 3 out of
12 is 220.
Challenge Problems
4.19 The sundae bar at Sarah’s favorite restaurant
has 5 toppings: hot fudge, sprinkles, walnuts,
cherries and whipped cream. In how many different
ways can Sarah top her sundae if she is restricted to
at most 2 toppings?
Use casework, based on the # of toppings Sarah
chooses. There are a total of 5 toppings, and her #
of toppings can be 0, 1, or 2. The order of the
toppings does not matter, so each case is counted by
a combination. The answer is
5
0
+
5
1
+
5
2
= 16
Problem 4.20
The Aviation Club at Rocco’s school has 12
members, including Rocco. They need to choose a
3-person Helicopter Committee and a 4-person
Glider Committee. Students can serve on either or
both committees, but Rocco refuses to serve on
both – he will only serve on one or the other or
neither. In how many ways can both committees be
chosen?
Approach the problem using complementary
counting. Ignore the restriction regarding Rocco,
then the total # of committees is 12 12 since there
3
are
12
4
ways to choose 3 people (out of 12) for Hel.
3
Com. and 12 ways to choose 3 people (out of 12)
4
from the Glider Com., and the choices are independent.
Now subtract the count of what we don’t want.
The total # of committees with Rocco serving on
both committees is 11 11 , because we choose the
3 4
2 others members for the Heli. Com. (from the 11
remaining club members) and choose the 3 other
members for the Glider Com. (from the 11 remaining club members). Therefore the final answer is
12 12
11 11
= 99,825.
3 4
2 3
Problem 4.21
There are 2 Senators from each of the 50 states. We
wish to make a a 3-Senator committee in which no
two members are from the same state.
(a) How many ways can we choose 3 states to be
represented?
(b) How many ways can we choose a Senator from
a chosen state?
(c) How many ways can the 3-Senator committee
be formed such that no 2 Senators are from the
same state?
There are 2 Senators from each of the 50 states. We
wish to make a a 3-Senator committee in which no
two members are from the same state.
(a) How many ways can we choose 3 states to be
represented? The order of the states does not matter
so this is a combination. 50C3 = 19,600.
There are 2 Senators from each of the 50 states. We
wish to make a a 3-Senator committee in which no
two members are from the same state.
(b) How many ways can we choose a Senator from
a chosen state? We want to choose 1 of the 2
Senators. The answer is 2C1 = 2.
There are 2 Senators from each of the 50 states. We
wish to make a a 3-Senator committee in which no
two members are from the same state.
(c) How many ways can the 3-Senator committee
be formed such that no 2 Senators are from the
same state? 1st we choose 3 states from 50, then
for each state we choose 1 of 2 Senators. Thus the
answer is
19,600 x 23 = 156,800.
Problem 4.22
There are 5 different pairs of gloves, where right &
left are distinguishable. Select 4 from 10 gloves.
(a)How many ways are there to select 2 pairs of
gloves?
(b) How many ways are there to select 4 such that
some 2 of the 4 make a pair?
Problem 4.22
There are 5 different pairs of gloves, where right &
left are distinguishable. Select 4 from 10 gloves.
(a)How many ways are there to select 2 pairs of
gloves?
There are 5 pairs and we need to select 2
without regard to order, so there are 5C2 = 10.
Problem 4.22
There are 5 different pairs of gloves, where right &
left are distinguishable. Select 4 from 10 gloves.
(b) How many ways are there to select 4 such that
some 2 of the 4 make a pair? It is easier to count
the ways to fail – namely, the # of ways to select 4
gloves so that we don’t have a pair. we have to take
one glove from each of 4 different pairs. There 5C4
= 5 choices for 4 pairs of gloves, then 2 choices for
the gloves within each pair (left or right), for a total
of 5 x 24 = 80 choices.
Problem 4.22
There are 5 different pairs of gloves, where right &
left are distinguishable. Select 4 from 10 gloves.
(b) How many ways are there to select 4 such that
some 2 of the 4 make a pair?
Without caring about the restriction, there are 10C4=
210 ways to choose any 4 gloves. Therefore, the #
of ways to choose 4 gloves such that we get at least
one pair is
210 – 80 = 130.
Problem 4.23
In poker, a 5-card hand is called three of a kind if
there are three cards of one rank and two other
cards which are not the same rank as each other or
as the other three cards. How many 5-card hands
are three of a kind? (Assume a standard 52-card
deck with 13 ranks in each of 4 suits.)
The 3 cards of the same rank can be any of the 13
ranks, and the suits have 4C3 possible combinations.
The next card can be any of 12 remaining ranks and
any of the 4 suits. The last card can be any of the 11
remaining ranks and any of the 4 suits. However,
the last 2 cards are interchangeable (their order
doesn’t matter), so divide by 2! to correct for overcounting. Thus the # of 3-of-a-kind hands is
(13 x 4) x (12 x 4) x (11 x 4) = 54,912
2!
Problem 4.24
Use combinations to find the number of distinct
arrangements of the letters of ‘ONONONONO’.
Let’s look at where the N’s will be. Once the
position of the N’s is determined, the arrangement
of the word is determined. Pick 4 positions for the
N’s out of 9. Since the N’s are not distinguishable
from each other, this is a combination.
9 = 126.
C
=
9 4
4
Problem 4.25
Let ABCDEFGH be a cube.
(a)How many different line segments can be formed
by connecting the vertices of the cube?
(b) How many different triangles can be formed by
connecting 3 of the vertices of the cube?
(c) How many non-congruent triangels can be
formed by connecting 3 of the vertices of the cube?
Problem 4.25
Let ABCDEFGH be a cube.
(a)How many different line segments can be formed
by connecting the vertices of the cube?
There are 8 vertices. The order of endpoints for a
segment does not matter, so it is a combination .
The # of segments is 8C2 = 28.
Problem 4.25
Let ABCDEFGH be a cube.
(b) How many different triangles can be formed by
connecting 3 of the vertices of the cube?
No 3 vertices are collinear, so any combination of 3
vertices will make a triangle.
Choosing 3 out of 8 is 8C3 = 56.
Problem 4.25
Let ABCDEFGH be a cube.
(c) How many non-congruent triangles can be
formed by connecting 3 of the vertices of the
cube?
Let the length of a side of the cube be 1.
A
1st, consider a ∆ that does not have an
B
edge of the cube as any of its 3 sides.
F
Then no 2 vertices of the ∆ can be on
the same edge of the cube. ∆ ACF has side
lengths of √2, √2, √2.
D
C
H
G
Problem 4.25
Next, if exactly one side of the ∆ is an edge
of the cube, then we can see that ∆ ADF is
an example of such a ∆, and all such ∆’s
have sides of length 1, √2, √3.
A
B
D
C
H
F
G
Problem 4.25
Finally, a ∆ cannot have all 3 of its sides
be edges of the cube, so the only case left is
where exactly 2 sides of the ∆ are edges of
the cube. An example is ∆ ABC.
Its sides are 1, 1, √2.
B
So there are 3 non-congruent ∆’s.
A
D
C
H
F
G