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Counting From Zero to mastery
By Sofian Makhlouf
Contents
Counting From Zero to mastery ....................................... 1
1. Methods and warm-up for counting. ........................ 2
1.1 How to find the cardinality of a finite set? ....... 2
1.2 How to find the cardinality of consecutive
terms? 3
1.3 How to use the multiplication rule of counting?
4
1.4 How to use the permutation of finite set? ........ 6
1.5 How to use the 𝒌 permutation of 𝒏 different
elements? ...................................................................... 7
1.6 How to use the permutation of 𝒏 distinct
elements with constraints? ........................................... 8
1.7 How to use the 𝒌 combination from 𝒏 different
elements? .................................................................... 10
1.8 How to place 𝒌 identical objects in 𝒏 different
places? ......................................................................... 12
1.9 How to use Pascal's identity? .......................... 14
1.10 How to use the binomial formula? .................. 16
2. Problems .................................................................. 17 3. Solution of problems ............................................... 19
1. Methods and warm-up for counting.
1.1 How to find the cardinality of a finite set?
M1) The cardinality of a set 𝐴 is a measure of the
"number of elements" of the set,
is denoted by 𝑐𝑎𝑟𝑑
or |𝐴| or #𝐴.
M2) |∅| = 0 and
N
M3) If 𝐴 is a subset of a finite set 𝑆, then 𝐴 = 𝑆 ∖ 𝐴
is the absolute complement of 𝐴 is the set of
elements in 𝑆 that are not in 𝐴. We have
|𝐴| = |S ∖ A | = |S| − |A|
M4)
M5) If 𝐴 ∩ 𝐵 = ∅ then |𝐴 ∪ 𝐵| = |𝐴| + |𝐵|
Warm-up 1
- Give the cardinality if each set:
𝐴 ∩ 𝐵 and 𝐴 = 𝑆 ∖ 𝐴
- In a class there are 100 students, each student
chooses at least one of two options 𝐴 and 𝐵. 60
students choose option 𝐴 and 70 students
choose option 𝐵. How many students choose
both options together?
Solution of warm-up 1
Solution of warm-up 2
- (a) 5 − 1 + 1 = 5 terms.
(b) 5 − 0 + 1 = 6 terms.
(c) 100 − 10 + 1 = 91 terms.
𝑛
𝑚
So: 𝑛 − 𝑚 =
𝑈
𝑛
−𝑈
𝑚
𝑎
. So, the cardinality of
consecutive terms from 𝑚 −th term to 𝑛 −th
term is: 𝑛 − 𝑚 + 1 =
𝑈
𝑛
−𝑈
𝑚
𝑎
- (a) The common difference is 𝑎 = 2 , so we have
10 − 2
2
(b) The common difference is 𝑎 = 2 , so we have
100 − 12
2
(c) The common difference is 𝑎 = 5 , so we have
118 − 13
5
(b) The common difference is 𝑎 = − 4 , so we
have
− 81 − 7
− 4
1.3 How to use the multiplication rule of counting?
M1) If a choice consists of 𝑘 steps, where the 𝑖 −th
step is done in 𝑛
𝑖
ways, then we have 𝑛
1
× … × 𝑛
𝑘
choices.
Warm-up 3
- With 1, 2 and 3 , how many 3-digit numbers
without repetition, can we have?
- With 1, 2 and 3 , how many 2-digit numbers with
repetition, can we have?
- How many ways we can build a password with two
letters followed by three digits? (With repetition).
- How many ways we can build a password with two
letters followed by three digits? (Without
repetition).
Solution of warm-up 3
- For the first number we have 3 choices,
for the second number we have 2 choices,
for the third number we have 1 choice.
So, we have 3 × 2 × 1 = 6 choices.
Those 3 - digit numbers are:
- For the first number we have 3 choices,
for the second number we have 3 choices,
for the third number we have 3 choices.
So, we have 3 × 3 × 3 = 27 choices.
Those 3 - digit numbers are:
- For the first letter we have 26 choices,
(5) In how many ways can six people sit in a row of six
chairs?
(6) A race has five horses. In how many orders can they
finish if they all finish at different times?
Solution of warm-up 4
(1) (a) 𝑃
3
𝑎
3
= 3! = 3 × 2 × 1 = 6 permutations.
(b) 𝛺 = {
( 1 , 2 , 3
) ,
( 1 , 3 , 2
) ,
( 2 , 1 , 3
) ,
( 2 , 3 , 1
) ,
( 3 , 1 , 2
) ,
( 3 , 2 , 1
) }
(2) 2! = 2 × 1 = 2 3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
5! = 5 × 4 × 3 × 2 × 1 = 120.
5!
3!
5 × 4 × 3 × 2 × 1
3 × 2 × 1
= 5 × 4 = 20
10!
7!
10 × 9 × 8 ×( 7 !)
7!
= 10 × 9 × 8 = 720
(4) There are 8! = 40 320 anagrams of word counting.
(5) There are 6! = 720 ways.
(6) There are 5! = 120 orders.
1.5 How to use the 𝒌 permutation of 𝒏 different
elements?
M1) The number of permutations of 𝑘 elements from
𝑛 distinct element, is:
𝑛
𝑎
𝑘
Warm-up 5
(1) How many ways we can build a password with three
digits? (Without repetition).
(2) How many ways we can build a password with
three letters? (Without repetition).
(3) A taekwondo tournament with 16 competitors. How
many ways can gold, silver and bronze medals be
awarded?
Solution of warm-up 5
(1) A password with three digits, without repetition is a
permutation of 3 elements from 10 distinct
elements, So, their number is 𝑃
10
𝑎
3
= 10 × 9 × 8
(2) A password with three letters, without repetition is
a permutation of 3 elements from 26 distinct
elements,
So, their number is 𝑃
26
𝑎
3
= 26 × 25 × 24 = 15600.
(3) There are 𝑃
16
𝑎
3
= 16 × 15 × 14 = 3360 ways.
1.6 How to use the permutation of 𝒏 distinct elements
with constraints?
M1) The number of permutations of set with 𝑛
distinct element, with constraint: “ 𝑘 elements
together” is:
𝑘! ×
- 𝑘 elements together are considered as one
object we arrange them in 𝑘! arrangements.
arrangements.
(2) The four books are considered as one object we
arrange them in 4! = 24 arrangements.
= 7 objects in 7! arrangements.
So, we have 4! × 7! = 24 × 5040 = 120960 ways
that the ten books be arranged, with the constraint
“books by the same author stay together”.
(3) Three Tunisians are considered as one object we
arrange them in 3! = 6 arrangements.
Four Canadians are considered as one object we
arrange them in 4! = 24 arrangements.
- 1 + 1 ) = 5 objects in 5! = 120
arrangements.
Therefore, we have 3! × 4! × 5! = 17280 ways
they can sit with the constraint: “the three Tunisians
are to sit together, and the four Canadians are to sit
together”.
1.7 How to use the 𝒌 combination from 𝒏 different
elements?
M1) A combination of n elements taken k at the time,
is a k-elements subset of an n-elements set.
their number is: (
𝑛
𝑘
𝑛!
𝑘!
( 𝑛−𝑘
) !
Warm-up 7
- Let 𝑆 = { 1 , 2 , 3 , 4 }
(a) How many subsets of 2 element of 𝑆 are there?
(b) Give Ω the set of all subsets of 2 element of S.
- Show that (
𝑛
𝑘
𝑛
𝑛−𝑘
3 ) Show that (
𝑛
0
4 ) Show that (
𝑛
1
5 ) Show that (
𝑛
2
𝑛
( 𝑛− 1
)
2
6 ) evaluate (
100
99
) and (
100
98
7 ) Give Ω the set of all subsets of 𝑆 = { 1 , 2 , 3 }. Give
- By giving the cardinality of all subsets of
𝑆 = { 1 , 2 , … , 𝑛} with 2 methods, show that:
𝑛
𝑘= 0
𝑛
Solution of warm-up 7
- (a) The number of subsets of 2 element of 𝑆, is
(b) The set of all subsets of 2 element of 𝑆 is:
𝑛
𝑛−𝑘
𝑛!
(𝑛−𝑘)!(𝑛−(𝑛−𝑘))!
𝑛!
𝑘!(𝑛−𝑘)!
𝑛
𝑘
𝑛
0
𝑛!
0 !(𝑛− 0 )!
𝑛!
𝑛!
M2) Let 𝒏 = 𝒏
𝟏
𝒌
and S a set with 𝑛
elements, where for all 𝑖 ∈ { 1 , … , 𝑘} , 𝑛
𝑖
elements are
identical.
The number of permutations of S is:
1
𝑘
M 3 ) Let 𝑛 = 𝑛
1
𝑘
and S a set with 𝑛 elements.
The number of possible divisions of 𝑆 into 𝑘 distinct
groups of respective sizes 𝑛
1
𝑘
is:
1
𝑘
Warm-up 8
- A coin is tossed 10 times. we get exactly 2 heads.
what is the number of possible cases?
- How many ways can you serve 99 cups of orange
juice and 2 cups of coffee for 101 guests?
- What is the number of anagrams of the word
BATATA?
- What is the number of anagrams of the word
BOSSBOOSSS?
- 6 children are to be divided into an A team and a B
team of 3 each. How many different divisions are
possible?
Solution of warm-up 8
- First method:
We want to place 2 identical objects in 10 different
places then we have: (
10
2
10 × 9
2
= 45 cases.
Second method:
We have 𝑛
1
= 2 for heads and 𝑛
2
= 8 for teals.
Thus, the number of ways is:
10!
2! 8!
- First method:
We want to place 99 identical objects in 101
different places then we have:
101
99
101
2
101 × 100
2
= 5050 ways.
Second method:
We have 𝑛
1
= 99 for cups of orange juice and
2
= 2 for cups of coffee.
Thus, the number of ways is:
101!
99! 2!
- We have 𝑛
1
= 1 for the letter B, 𝑛
2
= 2 for the
letters T and 𝑛
3
= 3 for the letter A: Thus, the
number of anagrams is:
5!
1! 2! 3!
- We have 𝑛 1
= 2 for the letter B, 𝑛
2
= 3 for the
letters O and 𝑛
3
= 5 for the letter S: Thus, the
number of anagrams is:
10!
2! 3! 5!
- The number of possible divisions of 6 children into 2
distinct groups A and B is:
6!
3! 3!
1.9 How to use Pascal's identity?
M1) (
𝑛+ 1
𝑘
𝑛
𝑘− 1
𝑛
𝑘
M2) Pascal’s triangle for 𝑛 = 10
- We have (
5
3
5
2
5 × 4
2
So (
6
3
5
2
5
3
We have (
6
2
6 × 5
2
So (
7
3
6
2
6
3
1.10 How to use the binomial formula?
M1) The binomial formula :
𝑛
𝑛
𝑘
𝑘
𝑛 𝑛−𝑘
𝑘= 0
M2) The coefficients in the binomial formula are the
entries of the last row of Pascal's triangle for 𝑛.
Warm-up 9
- Develop
3
- Give the coefficient of 𝑎
2
10
in
12
- Evaluate
𝑛
𝑘
𝑛
𝑘= 0
- Show that 𝑘(
𝑛
𝑘
𝑛− 1
𝑘− 1
Deduce that: ∑ 𝑘(
𝑛
𝑘
𝑘
𝑛 𝑛−𝑘
𝑘= 0
𝑛− 1
Solution of warm-up 9
- The binomial formula and Pascal’s triangle give:
3
3
0
0
3
3
1
1
2
3
2
2
1
3
3
3
0
3
2
2
3
- The coefficient of 𝑎
2
10
in
12
is:
12
2
12 × 11
2
𝑛
𝑘
𝑛
𝑘= 0
𝑛
𝑘
𝑘
𝑛−𝑘
𝑛
𝑘= 0
𝑛
𝑛
𝑛
𝑘
𝑛!
𝑘!
( 𝑛−𝑘
) !
( 𝑛− 1
) !
( 𝑘− 1
) !(𝑛− 1 −
( 𝑘− 1
) )!
𝑛− 1
𝑘− 1
Thus:
𝑛
𝑘
𝑘
𝑛−𝑘
𝑛
𝑘= 0
𝑛− 1
𝑘− 1
𝑘
𝑛−𝑘
𝑛
𝑘= 1
𝑛− 1
𝑘− 1
𝑘− 1
𝑛− 1 −
( 𝑘− 1
) 𝑛
𝑘= 1
𝑛− 1
𝑘− 1
𝑘
𝑛− 1 𝑛− 1 −𝑘
𝑘= 0
So
𝑛
𝑘
𝑘
𝑛 𝑛−𝑘
𝑘= 0
𝑛− 1
2. Problems
(1) Five people arranged in a row to be served at the
counter. How many possible ways are there?
(2) in computing, a bit is one of the integers { 0 , 1 } and a
word is any 32-bit string. How many possible words
are there?
(3) How many possible license plates are there with three
numbers and two letters? (With repetition)
(a) if the numbers come before the letters?
(b) if there is no restriction on where the letters and
numbers appear?
(4) How many possible license plates are there with three
numbers and two letters? (With no repetition)
(a) if the numbers come before the letters?
(b) if there is no restriction on where the letters and
numbers appear?
(5) A jar contains 10 balls numbered from 1 to 10.
Three balls are randomly drawn from the jar.
Find the number of draws in each case:
𝐴 = {𝑎, 𝑏, 𝑐, 𝑑} to 𝐵 = { 1 , 2 , 3 , 4 , 5 , 6 } such that
𝑓(𝑎 ) ≠ 1 and 𝑓
(15) Give the coefficient of 𝑥
3
in
5
(16) Give the coefficient of 𝑥
− 6
in
− 2
5
3. Solution of problems
(1) Each arrangement of 5 people is a permutation of 5
elements from 5 distinct elements,
So, their number is 𝑃 5
𝑎
5
= 5 × 4 × 3 × 2 × 1 = 120
(2) Two choices for each bit, and in a word, we have 32
bits. So, the number of possible words is: 2
32
(a) If the numbers come before the letters, then the
number of possible license plates is:
5
× 26
2
(b) If there is no restriction, then the number of
possible license plates is:
) × 10
5
× 26
2
Where (
5
2
) is the number of possible places of the
two letters.
(4)
(a) If the numbers come before the letters, then the
number of possible license plates (With no
repetition) is:
10 × 9 × 8 × 26 × 25 = 468000
(b) If there is no restriction, then the number of possible
license plates (With no repetition) is:
(
5
2
) × 10 × 9 × 8 × 26 × 25 = 4860000
Where (
5
2
) is the number of possible places of the
two letters.
(a) The draws are successive with repetition, then
the number of possible draws is:
10 × 10 × 10 = 1000
(b) The draws are successive without repetition,
then the number of possible draws is:
10 × 9 × 8 = 720
(c) In the case of simultaneous draws, the number
of draws is calculated as the number of
combinations of 3 balls drawn from 10.
Therefore, the total number of draws is:
10
3
10!
3!
( 10 − 3
) !
(6) 2 people must be chosen from 5, so we have:
5
2
5 × 4
2
= 10 choices.
(7) 2 men must be chosen from 5 and 1 woman must be
chosen from 4 , so we have:
5
2
×
4
1
= 40 choices.
(8) 3 people must be chosen from 5 men and 4 women
with at least 1 woman. So, we have:
5
2
×
4
1
5
1
×
4
2
5
0
×
4
3
= 74 choices
(9) For the first row we choose 3 men from 6 and 2
women from 4. There are (
6
3
) × (
4
2
) = 120 choices.
