Mathematics is a passion for me. Since I was
a kid, I have always loved doing some calculations. While I am getting older
day by day, my passion becomes much more than before. I hope that my passion
will keep increasing.
As I have started high school the topics
that we learn in mathematics lessons are getting harder. Although the topics
are harder than before, they become much more interesting. Therefore, I wanted
to do a research about the applications of permutation, combination and
binomial expansion in daily life.
While doing the research, I have learned
lots of new things about permutation and combination. I understood that there
are lots of interesting examples about this topic. Actually, we always select
and arrange some simple things, but we do not know that this concerns
permutation and combination.
By doing this project, I have recognized
that mathematics is our life.
Suzan R. Hofstede, 2021
Permutation of a set is the arrangement of
its members or objects. The elements of that set can be arranged in a sequence,
in a linear order or it can be rearranged if it was arranged before. This set
should be finite in order to do some calculations about it.
The formula is: r ≤ n, P(n,r) = n!/(n-r)!
In combination, one will only select the
elements of that finite set. The order of the selection does not matter. The
definition of combination is the number of possible arrangements in a set where
the order of the selection is not important.
The formula of combination is:
[1]. The formula of combination differs from the formula of permutation only a little bit. The only difference is the “k!” on the denominator part.
The binomial theorem describes how to expand
the algebraic expressions’ powers of a binomial. By using this teorem,
polinomials such as (x+y)n can be expanded easily.
While searching about this topic, I
understood that we use permutation, combination and binomial expansion in our
daily life, even if we do not know that is permutation, combination or binomial
theorem.
I have always loved mathematics since I was
a little kid. In my opinion, mathematics is in everywhere in our daily life.
The simpliest example is, we all use mathematics after we have finished our
shopping.
In this project my aim is to learn about
permutation, combination and binomial expansion in a
more detailed way. While searching about
this topic, I will also understand that we use permutation, combination and
binomial expansion in our daily life, even if we do not know that it is
permutation or combination.
The most basic example to permutation and
combination is about choosing our clothes, which is a very serious and
important topic for all girls in the world. For instance, everyday I think
about “what can I wear today”. First of all, I eliminate some of my clothes
according to the weather. After that I choose something according to my mood:
“Do I want to wear a skirt, a dress, or a pair thights?” All in all,
permutation and combination have a great importance in our lives.
While doing this project I will mainly
analyze what is permutation, what is combination and
what is binomial expansion at first. After
that I am going to make a research and analyze about “what can be the daily
life applications and examples of them.” I hope I will learn lots of new
information about this topic.
Permutation of a set is the arrangement of
its members or objects. The elements of that set can be arranged in a sequence,
in a linear order or it can be rearranged if it has been arranged before. This
set should be finite in order to do some calculations about it.
The formula is: r ≤ n, P(n,r) = n!/(n-r)!
For example, there is three people whose
names are Sophia, Blair and Cassandra. They can sit in 6 different arrangments.
We found the number 6 from “3!” which means three factorial. Factorial is a
symbol that we use for while we are going to multiply all the numbers before
that number. We also multiply that number as well.
If we want to use the formula of permutation,
firstly we should place the number “3” to the letter “n”. After that, as there
are “3” people who are going to be arranged, we should extract number “3” from
number “3”. Do not worry about the number zero, which is written in the place
where the denominators are written. We do not write only “0” to the
denominator, we write “0!” which is equal to “1”. The formular image is like
this: P(3,3) = 3!/(3-3)! = 3! = 6
The arrangements are:
1- Sophia –
Blair – Cassandra
2- Sophia –
Cassandra – Blair
3- Blair –
Sophia – Cassandra
4- Blair –
Cassandra – Sophia
5- Cassandra –
Sophia – Blair
6- Cassandra –
Blair – Sophia
The main difference between permutation and
combination is: in permutation you arrange the elements while in combination
you only select the elements of a set.
There are lots of examples from our daily
lives that we use permutation, but we really do not know that we are doing the
process of permutation. For instance, when we go to a theatre or a cinema we
sit in an order. But how do we determine that order? Especially if two people
or more want to sit together, or if a number of people do not want to sit
together; then there is a problem. We always solve this kind of problems
without being aware of what we are actually doing.
Repetitive permutation is different from
the permutation that we have talked about before. Repetitive permutation has a
different formula. If “n” is the number of elements, and “r” is the repetitive
element in the set, then: r ≤ n, P(n,r) = n!/r!
For instance, if we want to create
different numbers from the set {1, 1, 1, 2, 3, 4, 4}, then we should use the
formula. n = 7, r1 = 3 and r2 = 2. From 7!/3!x2! the
answer is 1680. 1680 different numbers can be created.
DAILY LIFE APPLICATIONS AND EXAMPLES OF PERMUTATION
1- Everyone has
a different T.R. identification number. This identification number is a number
which has 11 digits. The first digit is not zero, and the last digit is always
an even number. The last two digits are calculated according to the first 9
digits. Accordingly, how many T.R. identification number can be generated?
Since the first digit cannot be
"0", one can select only 9 numbers. Any of the 10 numbers can be
selected for the 2nd, 3rd, 4th, 5th,
6th, 7th, 8th, and 9th digits.
Since the last digit will be an even number; The numbers 0, 2, 4, 6 and 8 can
come. Therefore, 5 numbers can be selected.
As a result, 9. 108. 5 = 28.32.59
T.R. identification numbers can be created.
2- A genetic
engineer is doing research on a living creature’s gene. He/She will change the
order of 20 pairs base groups to obtain a more efficient gene in research. All
the adenine bases must match thymine bases and guanine bases must match
cytosine bases. Since there are 7 guanines and 7 cytosines to match, how many
different gene sequences can he/she get? (All guanines, thymines, cytosines,
and adenines are identical.)
7 out of 20
paired base groups are guanine and cytosine pairs. 13 of them are adenine and
thymine pairs. Since the bases in each organic base pair are identical, one
should use the repeated permutation. “20! / 7! . 13! ” different gene sequences
can be obtained.
3- In how many different ways can 10 shoes be
arranged on a 3-storey shelf? (if r is the number of shoes, and n is the number
of shelves; one can arrange the shoes P (n + r-1, r) different ways.)
In this case, one can arrange the shoes
P(12, 10) = 12! / 2! different ways.
4- A student has 15 books in total. This
student’s sister has 5 books which is the same with this student. (There are
two of the same 5 books). According to this information, in how many different
ways can these 20 books be placed on a single-storey shelf, provided that they
are side by side?
Since there
are two of the same 5 books out of the 20 books in total, a single-storey
shelf: they can arrange them in 20! / 5! different ways.
5-
Five members of the family will play a word
game. The aim of the game is to create meaningful or meaningless words by
choosing at least one of the 8 vowel letters. These words must have 5 letters.
There are 29 letters in total. Since the letters cannot be repeated, how many
words can be formed?
Words with 5 letters can be grouped as
follows:
-
Words which contain 1 wovel: C(8, 1) .
C(21, 4) . 5!
-
Words which contain 2 wovels: C(8, 2) .
C(21, 3) . 5!
-
Words which contain 3 wovels: C(8, 3) .
C(21, 2) .5!
-
Words which contain 4 wovels: C(8, 4) .
C(21, 1) .5!
-
Words which contain 5 wovels: C(8, 5) . 5!
Total number
of words that can be created is:
5! [C(8, 1) . C(21, 4) + C(8, 2) . C(21, 3)
+ C(8, 3) . C(21, 2) + C(8, 4) . C(21, 1) + C(8, 5)]
6- A family
will go to the cinema. This family consists of a mother, a father, a child, an
uncle and the two cousins of the child. According to this,
a- How many
different ways can this family sit in a row of 6 people?
They can sit in 6! = 720 different ways.
b- Unfortunataley,
6 seats could not be allocated in a single row. Three of them should sit in the
front row, and the others should sit in the back row. How many different ways
can they sit?
They can sit in C(6, 3) . 3! + C(3, 3) . 3!
= 120 + 6 = 126 different ways.
c- The two
brothers will not sit side by side since they never get along. How many
different ways can they sit in a row of 6 people?
(All situations - Situations where two siblings sit
next to each other = Situations where two siblings do not sit side by side)
All situations: 6!
Situations where two siblings sit next to each other:
5! . 2!
Situations where two siblings do not sit side by side:
6! – (5! . 2!) = 720 – 240 = 480 different arrangements can be created.
d- How many
different seating plans can be created since the mother, father and uncle will
sit side by side?
4! . 3! = 24 . 6 = 144 different
arrangements can be created.
In daily life,
there are many other different conditions. For instance, while the two siblings
are not going to sit together, one of the siblings may want to sit with their
cousin. The mother and
father may also
want to sit together. This family who consists of six people, may also sit in a
row where can ten people sit. All in all, many people encounter this kind of
situations during daily life. While solving these problems, we use combination
and permutation.
We have learned that what permutation is.
Then, one will understand that the explanation of combination is much easier.
In permutation we have arranged the elements of a finite set. In combination,
we will only select the elements of that finite set. The order of the selection
does not matter. The definition of combination is the number of possible
arrangements in a set where the order of the selection is not important.
The formula of combination is:
. The formula of combination differs from the formula of permutation only a little bit. The only difference is the “k!” on the denominator part. This difference comes from the idea that we are only selecting and not arranging. For instance, if we are going to select 4 items from a finite set, then we should figure out that we can select in 4 different orders. As we are not arranging them, these different 4 items have no importance. Therefore, we should divide into 4!
DAILY LIFE APPLICATIONS AND EXAMPLES OF
COMBINATION
1-
A student
was injured during the break and must go to the infirmary. Her teacher will
choose a friend among 29 people and send her to the infirmary. However, the
person who will go with the injured student should not be one of her 4 closest
friends, otherwise these two students may chat for a long time and be late for
the lesson. In this case, how many different ways can the student who can help
go to the infirmary be selected?
Since 4 of
the 29 people cannot be selected, the teacher should choose the student among
25 people.
29-4 = 25
C (25,1) = 25
different selections can be made.
2-
100 people applied to a team. A total of 15
people will be chosen to the team and 30 people will be taken as backups.
However, 10 people out of 100 applicants do
not fit the height criteria, and 5 people
do not have a good command of the terms on the subject. Since there are these
people who do not fit the desired characteristics, in how many different ways
can the team and the backups be determined?
15 out of
100 people will not be eligible for the team because they do not fit the
criterias. Therefore, selections will be made among 85 people. 15 of the 85
people can be taken into the team, and 30 of the remaining 70 people can be
taken into the backup team.
C (85,15) +
C (70,30) different selection can be made.
3-
There are a total of 6 different routes
from Ankara to Istanbul. There are 3 roads between Antalya-Ankara and 2
different roads between Ankara and Istanbul. There is one road that is not
connected to Ankara. A person departing from Antalya wants to visit Ankara to
visit his/her sister and will go to meet with his/her
friend in Istanbul. Since this person will
only visit his/her sister once during the round trip, how many different round
trips can be made between Antalya and Istanbul?
While going
to İstanbul, in
C (3,1). C
(2,1) = 6 different ways he/she can go. Since he/she will not stop at Antalya
on his/her way back, he/she can select only the particular 1 way. This
situation may be visa versa.
7.2 = 14
different round trips.
4- A student
will take an exam. This exam consists of 25 questions and each question has 5
options. Maximum 3 questions’ answers can be the same in a row. In this case,
how many different forms can the student's answer key be?
Any of the 5 answers can be selected in the
first three questions in this exam. In the 4th question, maximum of
4 different options can be done, because same 3 answers cannot be made in a
row. In the other two questions after the 4th question, any of the 5
options can be selected.
Likewise, only 4 options may be selected in
the 7th question. All the five options can be selected in questions
8 and 9. This cycle will continue as this until the 25th question.
Therefore, 517.48 = 517.216
different answer keys can be created.
5-
There are 50
employees in a company. The boss of this company is not included in these 50
people. There are 17 people dealing with the technical problems in total, 12
people dealing with the creativity of the presentation, 14 people working with
the content of the presentation, 7 people working with the food and beverage
service which will be made during the presentation. According to this
information,
a-
The boss
will select a CEO, a CEO assistant among these 50 people. How many different
pairs of CEO and CEO assistans can be selected?
50 CEO’s, 49 CEO assistants can be selected. 50.49 = 2450 different
selections can be made.
b- A group will be selected for a presentation. In this
group, there will be 5 people who are dealing with the creativity of the
presentation, 8 people dealing with the technical problems, 9 people dealing
with the content of the presentation and 3 people serving food and beverage
will be selected. How many different choices can be made?
People dealing with the technical problems: C(17,8)
People dealing with the creativity of the presentation: C(12,5)
People dealing with the content of the presentation: C(14,9)
People serving the food and beverage: C(7,3)
C(17,8) +
C(12,5) + C(14,9) + C(7,3) different selections can be made.
6- A person wants to send 4 cargoes. There are 5 different shipping
companies it can send. According to this information,
a- In how many different ways can these cargoes be sent?
Each cargo can be
sent with 5 different companies. Therefore, there are 54 possible situations.
They can be sent in 625 different ways in total.
b- The cargo which was sent by this person will be sent back to this
person. However, this returning process will not be done by the same shipping
company. In how many different ways can all the cargoes be sent back, provided
that they are not returned with the same company?
This person can
send 54 different ways. The returning process can be done in 44
different ways. In this case, 54 + 44 = 625 + 256 = 881
different sending operations can be made.
7- In a school, weekend courses are arranged for the students. Mathematics,
English and volleyball courses are held between 8:00 a.m. - 10:00 a.m.
Basketball and football courses are given between 10:00 a.m. - 12:00 a.m.
Between 13:00 - 15:00 p.m. physics and literature lessons are given. A student
will go to 3 courses in total, from 8:00 a.m. until 15:00 p.m. How many
different courses can he/she choose since he/she can choose only one of the
courses which are given at the same time?
C(3, 1). C(2, 1).
C(2, 1) = 3.2.2 = 12 different selections can be made.
8- Food and drinks in a restaurant are as follows:
·
Meals: There are 5 types of
kebab, 7 types of wraps, 4 types of hamburgers and 2 types of pasta.
·
Salads: There are 5 types of
salads.
·
Beverages: There are 3 types
of sodas, water, milk and ayran.
·
Desserts: There are 2 desserts
which are made of milk, 3 types of cake and 4 types of cookies.
According to this
information,
a- Two people will go to this restaurant. Each person will have one meal,
one drink, one salad, and one dessert. How many different ways can these people
make their meal choices?
There are 5 + 7 +
4 + 2 meals. There are 3 + 3 drinks. There are 5 salads. There are 2 + 3 + 4
desserts.
Each person, C(18,
1). C(6, 1). C(5, 1). C(9, 1) = 18.6.5.9 = 4860 different choices. Since two
people go to this restaurant, 4860.2 = 9720 different selections can be made.
b-
A person only wants to eat
foods that are healthy. He/She will not drink sodas, eat wraps, hamburgers or
pasta. He/She can only eat desserts which are made from milk. Accordingly, how
many different food choices can this person make?
This person can
select one of the 5 types of kebab and one of the 5 salads. He/She can order
water, milk or ayran. He/She can order one of the 2 desserts which are made
from milk.
C(5, 1) . C(5, 1)
. C(3, 1) . C(2, 1) = 5.5.3.2 = 150 different selections can be made.
c- A person who goes to this
restaurant has a lactose intolerance. While choosing a meal, she/he can eat
non-dairy foods. How many different foods can he/she select?
This person
will not be able to drink milk and ayran. He/She will not be able to eat any deserts which are made from milk.
C(18, 1) .
C(5, 1) . C(4, 1) . C(7, 1) = 18.5.4.7 = 2520 different selections can be
made.
9- A menu will
be created in a seafood restaurant. There are 7 types of fish and these fish
can be served between bread or with different types of salads. Since there are
3 types of bread and 6 types of salads, how many different meals can be written
on the menu?
7 different types of fish can be used for each bread
and salad. There are 3 + 6 = 9 types of bread and salads in total. 79
different types of meals can be written to this menu.
10- Everyone should make choices while selecting their clothes in daily
life. A girl has 5 shirts for winter, 4 shirts for summer, 18 sweatshirts and 7
trousers in her wardrobe. In addition, 8 out of 12 dresses are for summer; and
5 out of 8 skirts are for winter. T-shirts will be accepted for summer,
trousers for winter. Once a dress is selected, no other clothes can be worn. 7
of the 18 shirts are pink. 2 of the 8 skirts are red. 1 of 7 pants is red. Any
of the clothes which are for summer will be chosen.
a- How many different combinations can be made?
One of 4 shirts
and one of 3 skirts, one of 8 dresses, one of 18 t-shirts and 3 skirts can be
chosen.
In this case,
C(8,1) + C(4,1) . C(3,1) + C(18,1) . C(3,1) = 74 different selections can be
made.
b- A skirt and a shirt will be chosen. How many different combinations can
be made?
One of the 9
shirts and one of the 8 skirts will be chosen.
In this case,
C(9,1) . C(8,1) = 72 different combinations can be made.
c- A pair of trousers will be chosen, any other cloth will be chosen except
a dress. How many different combinations can be made?
One of 7 trousers
and one of 9 shirts, one of 7 trousers and one of 18 t-shirts can be selected.
In this case, C(7,1) . C(9,1) + C(7,1) . C(18,1) = 63 + 126 = 189 different
combinations can be made.
d- A combination will be made in any color except a red outfit under a pink
t-shirt. How many different combinations can be made?
Since 7 of the 18
t-shirts are pink, one of 7 can be chosen. Since 2 of the 8 skirts are red, one
should choose from 6 skirts. Since 1 of the 7 trousers is red, one should
choose from 6 trousers.
In this case,
C(7,1) . C(8-2,1)
+ C(7,1) . C(7-1,1)
= C(7,1) . C(6,1)
+ C(7,1) . C(6,1)
= 42 + 42 = 84
different combinations can be made.
Besides these
examples, there are many different conditions about choosing and arranging our
clothes. While thinking about in how many different ways one can put their
clothes in the closet, again people use combination and permutation. For
instance, one can arrange their clothes by putting them in order of colour. In
this case, one will determine wheter they will start from the left side or the
rigth side of their closet.
Additionally, each
group of clothing can be arranged. For example, the skirts may be put in a
particular place and the shirts may be put in a different specific place in the
closet. One may also arrange only their T-shirts in order of colour. Each
person uses combination, while thinking about which group of clothing he/she is
going to put in a certain shelf.
Combination and permutation are also used while
forming student’s classes in the schools. Students are first separated
according to their ages. After that, they are divided according to their
elective art courses, elective language courses and the curriculum they have
chosen.
For example, while separating the 9th and
10th grade students in a school, this kind of criteria is applied.
Their elective art classes are music and art. Their elective language courses
are Spanish, German and French. The curricula that the students will choose are
IB and Turkish National Curriculum.
After these students are grouped according to their
ages, they should also be grouped as IB-Turkish National curriculum, music-art,
Spanish-German-French. Thereafter, the students who made the same choices from
these three different courses should be determined and the classes should be
created. In addition to all these criteria, the number of boys and girls should
also be almost equal in each class.
In conclusion, many criteria are considered while
determining the classes in schools. The selections are made accordingly. When
making this kind of choices, combination is used.
The binomial theorem describes how to
expand the algebraic expressions’ powers of a binomial. By using this teorem,
polinomials such as (x+y)n can be expanded easily.
[2]
GENERAL RULES AND INFORMATION ABOUT
BINOMIAL EXPANSION
In an expansion of (x+y)n or
(x-y)n which has descending powers of x, there are some basic rules.
·
In binomial expansions, there are n+1
terms.
·
Sum of the powers of x and y is equal to
“n” in each term.
·
The coeeficients of the first term and the
last term is equal, in other words they are symmetric. The coefficients of the
second term from the begining and the second term from the end is also equal.
Generally, the coefficients of the terms which are equidistant from the beginnig
and the end are equal, as there is formula of “C(n, r) = C(n, r-1)”.
For
instance: C(n, 0) = C(n, n) , C(n, 1) = C(n, n-1) , C(n, r) = C(n, n-r)
·
In order to find the sum of the
coefficients, replace “1” to both x and y.
·
In order to find the constant term, replace
“0” to both x and y.
·
When “n” is even, the binomial expansion
has a middle term. The middle term is:
C(2n, n). xn. yn
·
The rth term form the beginning
is C(n, r-1). xn-(r-1). yr-1
·
The rth term form the end is
C(n, r-1). xr-1. yn-(r-1)
PASCAL’S TRIANGLE
In order to build the Pascal’s triangle,
one should start with “1” at the top of the triangle. Then one should continue
placing the numbers below that number in a triangular pattern. Each number is
the addition of the two numbers above that number.
The triangle is symmetrical. The rule of
the symmetrical coefficients comes from this triangle. The sum of the
horizontal lines is equal to 2row number. Therefore, each line is 2
times more than the previous line.
DAILY LIFE
APPLICATIONS AND EXAMPLES OF BINOMIAL EXPANSION
There are many different areas where the
binomial theorem is used. It is mostly used in mathematics. Besides
mathematics, there are also different daily life applications of this theorem;
such as in biology, in statistics, in modern physics, in the design of
infrastracture, in the architecture industry and in computing areas.
Binomial theorem is generally used in
probability theorem. In many countries, the economy is analyzed with
possibilities and probabilities. While countries try to develop their economic
system, they analyze their profits and losses by using binomial theorem. Many
financial jobs use the binomial theorem. This theorem is also used in order to
make realistic predictions in economics.
Furthermore, while predicting the weather
forecast, or the natural disasters again the binomial theorem is used.
Additionally, one can solve some impossible equations in some scientific
research. While determining the areas of infrastructure, the binomial theorem
is used. Another daily life application of binomial theorem is in computing
areas; while distributing the IP addresses.
Consequently, there are many different
areas where the binomial theorem is used.
Mathematics is the life
itself. One can see mathematics in any kind of working area in real life. In
this project, I have analyzed what permutation, combination and binomial
expansion are. Thereafter, I have
searched and analyzed what are the daily life applications of permutation,
combination and binomial expansion.
While preparing this
project, I realized that mathematics has a great importance in our lives. Even
when people choose a meal in a restaurant, they use combination. While ordering
our clothes in the closet, we use both permutation and combination. One may
think that these selections have no relation between mathematics. However, we
use mathematics even in these kinds of basic selections.
In conclusion, mathematics
is in everywhere. Especially the topics permutation, combination and binomial
expansion.
REFERENCES
1.
Doğan Necla, Alpar Nilay, Candemir Deniz,
The Math Book 10, Ankara, Turkish Education Association Publications, 2020
2.
Koç Ali Oktay, Duran Ahmet, Uçakçıoğlu
Çakır Gamze, Çetin Serpil, Mathematics For High School 10, Ankara, Turkish
Education Association Publications, 2020
3.
Nafile İlker, 10. Sınıf Matematik Konu
Kitabı, Ankara, Yazıt Yayınları, 2020
4.
Vapur Ahmet, Pehlivangil Hamza, 10. Sınıf
Matematik Planlı Ders Föyü, Ankara, Eğitim Vadisi Yayınları, Mayıs 2020
5.
Nafile İlker, 10. Sınıf Matematik Soru
Kitabı, Ankara, Yazıt Yayınları, 2020
6.
Duş Muharrem, Modüler Piramit Sistemiyle
Matematik Konu Anlatımı ve Soru Çözümü Kitabı, İstanbul, Karekök Yayınları,
2019
7.
Ed. Ayık Gonca, Ortaöğretim 10. Sınıf
Matematik Ders Kitabı, İzmir, Çağlayan Matbaası, 2020
8.
Duş Muharrem, Rutin Olmayan Problemler TYT
Soru Kitabı, İstanbul, Karekök Yayınları, 2019
9.
https://betterexplained.com/articles/easy-permutations-and-combinations/ (Date of access: 13.11.2020)
10.
https://bilgiyelpazesi.com/egitim_ogretim/konu_anlatimli_dersler/matematik_dersi_ile_ilgili_konu_anlatimlar/binom_acilimi_ozellikleri.asp (Date of access: 13.11.2020)
11.
https://www.britannica.com/science/permutation (Date of access: 13.11.2020)
12. http://www.bumatematikozelders.com/altsayfa/binom.html (Date of access: 13.11.2020)
13.
https://byjus.com/maths/permutation/ (Date of access: 13.11.2020)
14.
https://byjus.com/jee/binomial-theorem/ (Date of access: 13.11.2020)
15.
https://www.calculatorsoup.com/calculators/discretemathematics/permutations.php (Date of access: 13.11.2020)
16.
https://www.careerbless.com/aptitude/qa/permutations_combinations.php (Date of access: 13.11.2020)
17.
https://www.cevapbizde.com/kombinasyon-ve-permutasyon-gunluk-hayatta-nerelerde-kullanilir/ (Date of access: 13.11.2020)
18.
https://corporatefinanceinstitute.com/resources/knowledge/other/combination/ (Date of access: 13.11.2020)
19.
https://courses.lumenlearning.com/finitemath1/chapter/probability-using-permutations-and-combinations/ (Date of access: 13.11.2020)
20.
https://courses.lumenlearning.com/boundless-algebra/chapter/the-binomial-theorem/
21.
https://www.darussafaka.k12.tr/wp-content/uploads/2014/04/PROJE-36-B%C4%B0NOM-A%C3%87ILIMININ-GENELLENMES%C4%B0-VE-UYGULAMALARI-KULEL%C4%B0-ASKER%C4%B0-L%C4%B0SES%C4%B0-II.pdf (Date of access: 13.11.2020)
22.
https://dergipark.org.tr/en/download/article-file/201322 (Date of access: 13.11.2020)
23.
https://dictionary.cambridge.org/dictionary/english/combination (Date of access: 13.11.2020)
24.
https://eodev.com/gorev/9735108 (Date of access: 13.11.2020)
25.
https://eodev.com/gorev/318687 (Date of access: 13.11.2020)
26.
https://forum.donanimhaber.com/binom-aciliminda-sabit-terimi-nasil-buluyorduk--59300063 (Date of access: 13.11.2020)
27.
https://forum.donanimhaber.com/binom-acilim-ortak-terim--48560514 (Date of access: 13.11.2020)
28.
https://www.frmtr.com/lise-bilgi-istekleri/710521-odevistek-pascal-ucgeni-ve-binom-acilimi.html (Date of access: 13.11.2020)
29. http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html (Date of access: 13.11.2020)
30. http://www.khanacademy.org.tr/matematik-video.asp?ID=1877 (Date of access: 13.11.2020)
31. https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:polynomials/x9e81a4f98389efdf:binomial/v/binomial-theorem#:~:text=The%20Binomial%20theorem%20tells%20us,the%20process%20is%20relatively%20fast (Date
of access: 13.11.2020)
32. https://www.matematiktutkusu.com/forum/cozumlu-matematik-sorulari/19443-binom-acilimi-sorulari-cozumleri.html (Date of access: 13.11.2020)
33. https://matkafasi.com/53990/sabit-terim (Date of access: 13.11.2020)
34.
https://www.mathsisfun.com/combinatorics/combinations-permutations.html (Date of access: 13.11.2020)
35.
https://www.mathsisfun.com/algebra/binomial-theorem.html (Date of access: 13.11.2020)
36.
https://www.mathplanet.com/education/algebra-2/discrete-mathematics-and-probability/permutations-and-combinations (Date of access: 13.11.2020)
37.
https://www.matematiktutkusu.com/forum/cozumlu-matematik-sorulari/9889-binom-acilimi-sorulari-cozumleri-5-adet.html (Date of access: 13.11.2020)
38. https://www.mathwords.com/p/permutation_formula.htm (Date of access: 13.11.2020)
39. https://www.mathsisfun.com/pascals-triangle.html#:~:text=One%20of%20the%20most%20interesting,directly%20above%20it%20added%20together. (Date
of access: 13.11.2020)
40.
https://www.mbacrystalball.com/blog/2015/09/25/permutations-and-combinations/ (Date of access: 13.11.2020)
41.
https://medium.com/i-math/combinations-permutations-fa7ac680f0ac (Date of access: 13.11.2020)
42.
https://www.merriam-webster.com/dictionary/combination (Date of access: 13.11.2020)
43. http://www.mynet.com/cevaplar/binom-kombinasyon-ve-permutasyon-nerelerde-kullanilir/6941710 (Date of access: 13.11.2020)
44.
https://www.onlinemathlearning.com/combinations.html (Date of access: 13.11.2020)
45. https://www.purplemath.com/modules/binomial.htm (Date of access: 13.11.2020)
46. https://stattrek.com/statistics/dictionary.aspx?definition=combination (Date of access: 13.11.2020)
47.
https://teknoseyir.com/blog/t-c-kimlik-numaralarinin-algoritmasi (Date of access: 13.11.2020)
48. https://www.turkhackteam.org/off-topic/1792392-permutasyon-ve-binom-aciliminin-kullanim-alanlari.html (Date of access: 13.11.2020)
49. https://www.quora.com/What-are-some-real-world-examples-of-the-use-of-the-binomial-theorem (Date
of access: 13.11.2020)
50. https://en.wikipedia.org/wiki/Permutation (Date of access: 13.11.2020)
51. https://en.wikipedia.org/wiki/Binomial_theorem (Date of access: 13.11.2020)
52. http://yildizlaranadolu.com/wp-content/uploads/2019/04/35-Binom-A%C3%A7%C4%B1l%C4%B1m%C4%B1.pdf (Date of access: 13.11.2020)
53.
https://www.youtube.com/watch?v=cvhyJT9c0ac (Date of access: 13.11.2020)
Yorumlar
Yorum Gönder