permutations
MBOU "Yanishevskaya Basic School"
Teacher: Zvereva T.I.
Solve the problem:
Anton, Boris and Victor bought
3 football tickets for 1st, 2nd, 3rd places in the first row of the stadium. In how many ways can the boys take these places?
The solution of the problem:
- There might be a sequence like this:
A B C A C B
May be so:
B C A B A C
Or maybe like this:
C A B C B A
Answer: 6 options
Remember!!!
permutation called a set of n elements recorded in a specific okay.
- The theorem on permutations of elements of a finite set:
n different elements can be arranged one at a time n different places exactly n! ways.
The number of ways is equal to the number of permutations
from 3 elements. By the formula for the number of permutations, we find that
Р3=3!= 1 ∙ 2 ∙3 = 6
Five friends decided to take a photo. In how many ways can this be done?
Task:
In the 9th grade on Wednesday there are 6 lessons: mathematics, literature,
Russian language, English language, biology and physical culture. How many scheduling options are available?
Let's put the items in order:
Total options
schedules:
1 ∙ 2∙ 3 ∙ 4 ∙5 ∙ 6 = 720
Thing
Mathematics
Number of options
Literature
Russian language
English language
Biology
Physical Education
Task:
- There are nine different books, four of which are textbooks. In how many ways can these books be arranged on the shelf so that all the textbooks are next to each other? Plan:
1) textbooks = book
2) R6 permutations of books
3) P6=6!
4) P4 permutation of textbooks
5) P4=4!
6) P 6 ∙ P4
Homework:
1. In spring, mother buys a lot of fruits for her child. She bought a banana, an apple, an orange, a lemon, a pear and a kiwi. Find the number of possible options for eating fruits.
2 . Eleven players line up before the start of the match. The first is the captain, the second -
goalkeeper and the rest randomly.
How many ways are there to build?
3. In how many ways can 10 books be arranged on a shelf, 4 of which are books by the same author and the rest by different authors, so that the books of the same author stand side by side?
Until we meet again
with combinatorial problems
Permutations are combinations made up of the same elements and differing in their order. The number of all possible permutations of elements is denoted by P n, and can be calculated by the formula: Permutation formula: P n =n! During permutation, the number of objects remains unchanged, only their order changes. As the number of objects increases, the number of permutations grows very quickly and it becomes difficult to depict them visually.
Task 1. Seven teams participate in the tournament. How many options for the distribution of seats between them are possible? Р 7 =7!=1*2*3*4*5*6*7=5040 Answer: 5040 Problem 2. In how many ways can 10 people sit at the round table? P 10 =10! = = 1*2*3*4*5*6*7*8*9*10 = Answer:
Let there be n different objects. We will choose m objects from them and rearrange them in all possible ways among themselves. The resulting combinations are called placements of n objects by m, and their number is equal to: When placing, both the composition of the selected objects and their order change. Placement formula:
Task: In how many ways can three vouchers to one sanatorium be distributed among five people? Since vouchers are provided to one sanatorium, the distribution options differ from each other for at least one person. Therefore, the number of ways to distribute Answer: 10 ways.
Task: 12 people work in the workshop: 5 women and 7 men. In how many ways can a team of 7 people be formed so that there are 3 women in it? Of the five women, three must be selected, so the number of selection methods. Since it is required to select four men out of seven, then the number of ways to select men Answer: 350
Fundamentals of combinatorics.
Placements, permutations,
combinations.
naughty monkey
Donkey,
Goat,
Yes, clubfoot Mishka
Started to play a quartet
Stop, brothers, stop! -
Monkey shouts, - wait!
How does the music go?
You don't sit like that...
And so, and so transplanted - again the music is not going well.
Here, more than ever, their analysis went
And disputes
Who and how to sit ...
know:
- definitions of the three most important concepts of combinatorics:
- placement of n elements by m;
- combinations of n elements by m;
- permutations of n elements;
- basic combinatorial formulas
- to distinguish tasks into "permutations", "combinations", "placements" from each other;
- apply the basic combinatorial formulas in solving the simplest combinatorial problems.
be able to:
a bunch of
A set is characterized by the union of some homogeneous objects into a single whole.
The objects that form a set are called elements of the set.
We will write the set by placing its elements in curly brackets ( a, b, c, … , e, f}.
In a set, the order of the elements does not matter, so ( a, b} = {b, a}.
A set that does not contain any element is called empty set and is denoted by the symbol ø.
a bunch of
If each element of the set BUT is an element of the set B, then we say that the set BUT is a subset of the set AT.
A bunch of ( a, b) is a subset of the set ( a, b, c, … , e, f}.
Denoted
List the possible options for a subset of the set ( 3 , 4 , 5 , 7, 9 }.
Combinatorics is a branch of mathematics that studies questions about how many different combinations subject to certain conditions can be made from elements belonging to a given set.
Combinatorics is an important branch of mathematics that studies the patterns of arrangement, ordering, selection and distribution of elements from a fixed set.
SUMMATION RULE
If two mutually exclusive actions can be performed according to k and m ways, then one of these actions can be performed k+m ways.
Example #1
From city A to city B can be reached by 12 trains, 3 planes, 23 buses. How many ways can you get from city A to city B?
Decision
Example #2
There are n colored balls in a box. Randomly take out one ball. In how many ways can this be done?
Decision. Certainly, n ways.
Now these n balls are distributed in two boxes: In the first m balls, in the second k. Randomly take out one ball from any box. In how many different ways can this be done?
Decision.
The ball can be drawn from the first box m in various ways, from the second k in various ways, total N = m + k ways.
PRODUCT RULE
Let two actions performed one after the other can be carried out in accordance k and m ways Then both of them can be done k ∙ m ways.
Example #3
8 hockey teams take part in the tournament. How many ways are there to distribute first, second, and third places?
Decision
Example #4
How many two-digit numbers can be written in decimal notation?
Decision. Since the number is two-digit, the number of tens ( m) can take one of nine values: 1,2,3,4,5,6,7,8,9. Number of units ( k) can take the same values and can, in addition, be equal to zero. Hence it follows that m= 9, and k= 10. In total we get two-digit numbers
N= m · k= 9 10 =90.
Example #5
There are 14 girls and 6 boys in the student group. In how many ways can two students of the same gender be chosen to complete different tasks?
Decision. According to the multiplication rule of two girls, you can choose 14 13 = 182 ways, and two boys 6 5 = 30 ways. Two students of the same gender should be selected: two students or female students. According to the addition rule of such choices,
N = 182 + 30 = 212.
Connection types
Sets of elements are called compounds.
There are three types of connections:
- permutations from n elements;
- accommodation from n elements by m;
- combinations of n elements by m (m < n).
Definition: permutation from n elements is any ordered set of n elements.
In other words, this is such a set for which it is indicated which element is in the first place, which one is in the second, which one is in the third, ..., which one is in the nth place.
PERMUTATIONS
Permutations are such connections n elements from given elements that differ from one another in the order of the elements.
The number of permutations of n elements is denoted by Pn.
Рn = n · ( n- one) · ( n– 2) · … · 2 · 1 = n!
Definition:
Let be n- natural number. Through n! (read "en factorial") denotes a number equal to the product of all natural numbers 1 from to n:
n! = 1 2 3 ... n.
If n= 0, by definition it is assumed: 0! = 1.
FACTORIAL
Example #6
Let's find the values of the following expressions: 1! 2! 3!
Example #7
What is equal to
a) R 5 ;
b) R 3.
Example #8
Simplify
b) 12! 13 14
in) κ ! · ( κ + 1)
Example #9
In how many ways can 8 participants in the final race be placed on eight treadmills?
Decision.
R 8=8! = 8 7 6 5 4 3 2 1 =40320
ACCOMMODATION
Definition. Accommodation from n elements by m any ordered set is called m elements, consisting of elements n element set.
Number of placements from m elements by n stand for:
calculated by the formula:
Example #9
Students of the 11th grade study 9 subjects. In the schedule of training sessions for one day, you can put 4 different subjects. How many different ways are there to schedule a day?
Decision.
We have a 9-element set, the elements of which are educational subjects. When scheduling, we will choose a 4-element subset (of lessons) and set the order in it. The number of such ways is equal to the number of placements out of nine by four ( m=9, n=4) i.e A 94:
Example #10
In how many ways can a prefect and an assistant prefect be chosen from a class of 24 students?
Decision.
We have a 24-element set, the elements of which are students of the class. In the election of the headman and assistant headman, we will choose a 2-element subset (student) and establish an order in it. The number of such ways is equal to the number of placements out of nine by four( m=24, n=2), i.e A 242:
COMBINATIONS
Definition. A combination without repetitions from n elements by m- called any m elemental subset n-element set
The number of combinations of n elements by m is denoted
and calculated by the formula:
Example #11
In how many ways can two attendants be chosen from a class of 24 students?
Decision.
n =24, m=2
COMBINATIONS
ACCOMMODATION
PERMUTATIONS
Рn = n!
Determine what type of connections the task belongs to.
1. In how many ways can you schedule one school day from 5 different lessons?
2. There are 12 students in class 9B. In how many ways can a team of 4 be formed to participate in the Mathematical Olympiad?
Is the order of the elements in the join taken into account?
Are all elements included in the connection?
Conclusion: permutation
Is the order of the elements in the join taken into account?
Are all elements included in the connection?
(this question does not need an answer)
Conclusion: combinations
3. How many different two-digit numbers are there, in which the numbers 1, 2, 3, 4, 5, 6 can be used, if the numbers in the number must be different?
Is the order of the elements in the join taken into account?
Are all elements included in the connection?
Conclusion: placement
naughty monkey
Yes, clubfoot Mishka
Started to play a quartet
Stop, brothers, stop! -
Monkey shouts, - wait!
How does the music go?
You don't sit like that...
And so, and so transplanted - again the music is not going well.
Here, more than ever, their analysis went
Who and how to sit ...
How many different arrangements of musicians are possible?
Decision.
Is the order of the elements in the join taken into account?
Are all elements included in the connection?
Conclusion: permutation
Рn = n! =n · ( n- one) · ( n– 2) · … · 2 · 1
P4 = 4! = 4 3 2 1=24
“Sooner or later every correct mathematical idea finds application in this or that business”?
permutations
accommodation
combination
Problem Solving Results
HOMEWORK
Learn abstract and formulas.
S. 321 No. 1062
COMBINATORICS
Lesson Objectives:
- Find out what combinatorics studies
- Find out how combinatorics originated
- Learn the formulas of combinatorics and learn how to apply them in solving problems
The birth of combinatorics as a branch of mathematics is associated with the works of Blaise Pascal and Pierre Fermat on the theory of gambling.
Blaise Pascal
Pierre Fermat
A great contribution to the development of combinatorial methods was made by G.V. Leibniz, J. Bernoulli and L. Euler.
G.V. Leibniz
L. Euler.
I. Bernoulli
Lemma. Let there be m elements in set A and n elements in set B. Then the number of all distinct pairs (a,b) where a\in A,b\in B will be equal to mn. Proof. Indeed, with one element from the set A, we can make n such different pairs, and in total there are m elements in the set A.
Placements, permutations, combinations Let's say we have a set of three elements a,b,c. In what ways can we choose two of these elements? ab,ac,bc,ba,ca,cb.
Permutations We will rearrange them in all possible ways (the number of objects remains unchanged, only their order changes). The resulting combinations are called permutations, and their number is PN = n! =1 · 2 · 3 · ... · ( n-1) n
Symbol n! is called factorial and denotes the product of all integers from 1 to n. By definition, they consider that 0!=1 1!=1 An example of all permutations of n=3 objects (different shapes) is in the picture. According to the formula, there should be exactly P3=3!=1⋅2⋅3=6 , and so it turns out.
With an increase in the number of objects, the number of permutations grows very quickly and it becomes difficult to depict them visually. For example, the number of permutations of 10 items is already 3628800 (more than 3 million!).
Accommodations Let there be n different objects. We will choose m objects from them and rearrange them in all possible ways among themselves (that is, both the composition of the selected objects and their order change). The resulting combinations are called placements of n objects by m, and their number is equal to Aⁿm =n!(n−m)!=n⋅(n−1)⋅...⋅(n−m+1)
Definition. By placing a set of n distinct elements over m elements (m ≤ n) called combinations , which are composed of given n elements by m elements and differ either in the elements themselves or in the order of the elements.
Combinations Let there be n different objects. We will select m objects from them in all possible ways (that is, the composition of the selected objects changes, but the order is not important). The resulting combinations are called combinations of n objects by m, and their number is equal to Cmn=n!(n−m)!⋅m!
An example of all combinations of n=3 objects (different shapes) by m=2- in the picture below. According to the formula, there must be exactly C23=3!(3−2)!⋅2!:3!=3. It is clear that there are always fewer combinations than placements (because the order is important for placements, but not for combinations), and it is precisely in m! times, that is, the relation formula is correct: Amn=Cmn⋅Pm.
Method 1. 2 people participate in one game, therefore, you need to calculate how many ways you can select 2 people out of 15, and the order in such pairs is not important. We use the formula to find the number of combinations (samples that differ only in composition) of n different elements by m elements
n!= 1⋅ 2 ⋅3⋅...⋅ n , for n=2, m=13.
Method 2. The first player played 14 games (with the 2nd, 3rd, 4th, and so on until the 15th), the 2nd player played 13 games (3rd, 4th, etc. until the 15th th, we exclude the fact that there was already a game with the first), 3rd player - 12 games, 4th - 11 games, 5 - 10 games, 6 - 9 games, 7 - 8 games, 8 - 7 games,
and the 15th already played with everyone.
Total: 14+13+12+11+10+9+8+7+6+5+4+3+2+1=105 games
ANSWER. 105 parties.
Mathematics teacher Aksenova Svetlana Valerievna
Bugrovskaya secondary school of the Vsevolozhsky district of the Leningrad region
slide 1
slide 2
slide 3
slide 4
slide 5
slide 6
Slide 7
Slide 8
Slide 9
Slide 10
slide 11
slide 12
slide 13
Slide 14
slide 15
slide 16
Slide 17
Slide 18
Slide 19
Slide 20
slide 21
slide 22
slide 23
slide 24
The presentation on the topic "Discrete analysis. Combinatorics. Permutations" can be downloaded absolutely free of charge on our website. Project subject: Informatics. Colorful slides and illustrations will help you keep your classmates or audience interested. To view the content, use the player, or if you want to download the report, click on the appropriate text under the player. The presentation contains 24 slide(s).
Presentation slides
slide 1
Discrete Analysis
Lecture 3 Combinatorics. Permutations
slide 2
Permutations
Let a set of n elements be given. The ordering of these elements is called a permutation. Sometimes "out of n elements" is added. Usually one, basic or natural, ordering is distinguished, which is called a trivial permutation. The elements of the set A are not of interest to us. Often, integers from 1 to n or from 0 to n-1 are taken as elements. Denote the set of permutations of n elements by Pn and its cardinality by Pn. Let's ask the same three questions: what is Pn equal to, how to iterate over the elements of Pn , how to renumber them.
slide 3
Theorem on the number of permutations
The number of permutations of n elements is n! - the product of numbers from 1 to n. (n! reads n-factorial) Proof. By induction. For n=1 the formula is obviously correct. Let it be true for n-1, we will prove that it is true for n as well. The first element of the ordering can be chosen in n ways, and the rest can be assigned to the chosen first element in ways. Therefore, the formula Pn= Pn-1n is correct. If Pn-1=(n-1)!, then Pn=n!
slide 4
Permutation numbering
To enumerate the permutations, we map the set Pn one-to-one to another set Tn, on which it will be much easier to enumerate, and then for any element pPn we take the number of its image in Tn as its number. The set Tn is a direct product of several numerical segments Tn =(0:(n-1))(0:(n-2) ... (0). That is, each element of Tn is a set of non-negative numbers i1, i2, …, in-1, in, and ikn-k.
slide 5
Display
Take a permutation and write down a trivial permutation next to it. As the first index, we take the place of the first element (counting from zero) in the trivial permutation. Instead of a trivial permutation, let's write a string of the remaining characters. As the second index, take the place of the second element of the permutation in this row. Let's repeat the process to the end. Obviously, the kth index will be less than the length of the kth row, and the last index will be equal to zero. Let's see this process with an example.
slide 6
Display example
0 1 2 3 4 5 6 Index c a d f g b e a b c d e f g 2 2 a d f g b e a b d e f g 0 2 0 d f g b e b d e f g 1 2 0 1 f g b e b e f g 2 2 0 1 2 g b e b e g 2 2 0 1 2 2 b e b e 2 0 2 0 2 e 2 It is obvious that this process is reversible and the inverse mapping will construct the initial permutation from the set of indices.
Slide 7
Numbering of the set Tn
Any direct product of ordered sets can be viewed as a number system with a variable base. Think back to the seconds example from the first lecture, or consider some old size scale: 1 yard = 3 feet, 1 foot = 12 inches, 1 inch = 12 lines, 1 line = 6 points. (2, 0, 4, 2, 3) = 2 yards 0 feet 4 inches 2 lines 3 points, how many points is that? You need to count (this is called Horner's scheme) (((2 3+0) 12+4) 12+2) 6+3
Slide 8
Numbering of the set Tn - 2
The formula # that finds the number for the set of indices i1, i2, ..., in-1, in, we prefer to write in the form of recursive expressions #(i1, i2, ..., in) = a(i1, i2, ..., in-1, n-1); a(i1, i2, …, ik,k) = a(i1, i2, …, ik-1,k-1)(n-k+1)+ ik; a(empty,0) = 0; According to this formula #(2,0,1,2,2,0,0) = a(2,0,1,2,2,0,6). We have a(2,1)=2; a(2,0,2) = 26+0=12; a(2,0,1,3)=125+1=61; a(2,0,1,2,4) =614+2=246; a(2,0,1,2,2,5) =2463+2=740; a(2,0,1,2,2,0,6) =7402+0=1480;
Slide 9
Iterating over sets of indices
Based on the foregoing, it is easy to enumerate permutations: you need to enumerate all sets of indices from and, for each set, build the corresponding permutation. To enumerate sets of indices, we note that in fact each set is a record of a number in a special number system with a variable base (the system is called factorial). The rules for adding 1 in this system are almost as simple as in binary: moving from the last bit, try to add 1 in the current bit. If possible, then add 1 to stop. If not possible, write bit zero and move on to the next bit.
Slide 10
Iterating over index sets - 2
Consider this example: 7 6 5 4 3 2 1 These are the variable bases 3 4 4 2 1 1 0 3 4 4 2 2 0 0 Note that in 3 4 4 2 2 1 0 each line begins with 3 4 4 3 0 0 0 the same as in the previous one, 3 4 4 3 0 1 0 then comes the element, strictly 3 4 4 3 1 0 0 greater, . . . , and 3 4 4 3 1 1 0 what follows is not essential. 3 4 4 3 2 0 0 Therefore, each line 3 4 4 3 2 1 0 is lexicographically greater than 3 5 0 0 0 0 0 of the previous one. 3 5 0 0 0 1 0
slide 11
Lexicographic enumeration of permutations theorem
The described algorithm enumerates the permutations in lexicographical ascending order. Proof. It is enough for us to show that if we have two sets of indices I1 and I2, and I1 lexicographically precedes I2, then the permutation (I1) lexicographically precedes (I2). These permutations are formed sequentially, and as long as I1 and I2 match, so do their permutations. A larger index value corresponds to a larger element.
slide 12
Direct algorithm for lexicographic enumeration of permutations
We take some permutation p and directly find the lexicographically next one. Let's take the beginning - the first k elements. Among its extensions, the minimum is known, in which all elements are arranged in ascending order, and the maximum, in which they are arranged in descending order. For example, in the permutation p =(4, 2, 1, 7, 3, 6, 5) all extensions for (4, 2, 1) lie between (3, 5, 6, 7) and (7, 6, 5, 3). The existing continuation is less than the maximum, and the 3rd element can still be left unchanged. And the 4th too. And the 5th one needs to be changed. To do this, from the remaining elements, you need to take the next one in order, put it on the 5th and assign the minimum continuation. It turns out (4, 2, 1, 7, 5, 3, 6).
slide 13
Direct algorithm for lexicographic enumeration of permutations - 2
Let us write down the following permutations. (4, 2, 1, 7, 5, 3, 6) (4, 2, 1, 7, 5, 6, 3) (4, 2, 1, 7, 6, 5, 3) (4, 2, 3, 1, 5, 6, 7) (4, 2, 3, 1, 5, 7, 6) (4, 2, 3, 1, 6, 5, 7) (4, 2, 3, 1, 6 , 7, 5) (4, 2, 3, 1, 7, 5, 6) (4, 2, 3, 1, 7, 6, 5) (4, 2, 3, 5, 1, 6, 7)
Slide 14
Formal description of the algorithm
Working state: Permutation p and boolean sign isActive. Initial state: A trivial permutation is written to and isActive = True. Standard step: If isActive, return the permutation as the result. Moving from the end, find the largest monotonically decreasing suffix in the permutation. Let k be the position before the suffix. Put isActive:= (k > 0). If isActive, then find the smallest element in the suffix that exceeds pk, swap it with pk, and then “flip” the suffix.
slide 15
Another permutation algorithm
Let us now try to enumerate the permutations so that two successive permutations differ little from each other. How little? By one elementary transposition in which two adjacent elements are swapped. Is it possible? We will show the schematic diagram of such an algorithm, it will be of interest to us. Imagine n-1 elementary "mechanisms", each of which moves its element within the set. At each step, the mechanism makes a shift to the left or right. The direction changes when the element reaches the edge. It takes one step to change direction, during which the next mechanism takes a step, which, however, can also change direction.
slide 16
Another permutation enumeration algorithm -2
Let's see an example. 1 2 3 4 5 Чей ход 1 2 3 4 5 Чей ход a b c d e a c d a b e a b a c d e a c d b a e a b c a d e a c d b e a b b c d a e a c d e b a a b c d e a b c d e a b a c b d e a a c d a e b a c b d a e a c a d e b a c b a d e a a c d e b c c a b d e a a d c e b a a c b d e b d a c e b a a c d b e a d c a e b a c a d b e a d c e a b a
Slide 17
List of permutations. one
function ExistsNextPerm(var kCh: integer): Boolean; var iV,iP,iVC,iPC: integer; beginresult:= False; for iV:= nV downto 2 do if count
Slide 18
The problem of the minimum sum of pairwise products
Let two sets of n numbers be given, say, (ak|k1:n) and (bk|k1:n) . These numbers are divided into pairs (ak,bk) and the sum of their pairwise products k1:n akbk is calculated. You can change the numbering (ak) and (bk). It is required to choose such numbering at which the sum is minimal. In this problem, one can fix some numberings (ak) and (bk) and look for a permutation for which the minimum of the sum k1:n akb(k) is attained. We will choose numberings when (ak) are in ascending order and (bk) are in descending order.
Slide 19
Theorem on the minimum sum of pairwise products
The minimum of the sum of pairwise products is achieved on a trivial permutation. Proof. Suppose that there are two indices k and r such that ak 0, i.e. ar br + ak bk > ar bk + ak br . In our numbering (ak) are arranged in ascending order. If (bk) are not in ascending order, then there is a pair of k and r, as mentioned above. By rearranging bk and br for this pair, we will reduce the value of the sum. Hence, in the optimal solution (bk) are in ascending order. This simple theorem will be encountered several times in what follows.
Slide 20
The maximum increasing subsequence problem
A sequence (ak|k1:n) of numbers of length n is given. It is required to find its sequence of the greatest length, in which the numbers (ak) would go in ascending order. For example, in the sequence 3, 2, 11, 14, 32, 16, 6, 17, 25, 13, 37, 19, 41, 12, 7, 9, the maximum subsequence will be 2, 11, 14, 16, 17, 25, 37, 41 This problem is related to permutations because the original sequence can be a permutation. We will limit ourselves to showing how this problem is solved, and we will leave the formalization and justification of the algorithm to the audience.
slide 21
Finding the maximum increasing subsequence
We will split ours as economically as possible into decreasing sequences (example modified) the topmost of the lines where it does not break the order. Let's take the number from the bottom row, 21. Why is it in the 8th row? 19 interferes with it. But 17 interferes with 19. And 16 interferes with it. And so on. The sequence 9, 11, 14, 16, 17, 19, 21 increases and has length 7. Any sequence of greater length contains two numbers from one line Dirichlet principle) and cannot be increasing.
slide 22
The problem of the minimum number of inversions
A sequence (ak|k1:n) of numbers of length n is given. An inversion is a mirror reflection of some of its substrings, a continuous subsequence, performed in place. It is required to arrange the elements of the sequence in ascending order for the minimum number of inversions. For example, a “solid” permutation can be transformed like this (the red letters are rearranged, the large ones are already in place)
slide 23
Exam questions
Permutations. Their enumeration and numbering. The problem of the minimum of the scalar product. The problem of the largest increasing subsequence.
slide 24
1. Two-way transition permutation number 2. Find a permutation that is a given number of steps away from the given one. 3. Enumerating permutations by elementary transpositions. 4. Run an example for the problem of the minimum of the scalar product.
Tips on how to make a good presentation or project report
- Try to involve the audience in the story, set up interaction with the audience using leading questions, the game part, do not be afraid to joke and smile sincerely (where appropriate).
- Try to explain the slide in your own words, add additional interesting facts, you don’t just need to read the information from the slides, the audience can read it themselves.
- No need to overload your project slides with text blocks, more illustrations and a minimum of text will better convey information and attract attention. Only the key information should be on the slide, the rest is better to tell the audience orally.
- The text must be well readable, otherwise the audience will not be able to see the information provided, will be greatly distracted from the story, trying to make out at least something, or completely lose all interest. To do this, you need to choose the right font, taking into account where and how the presentation will be broadcast, and also choose the right combination of background and text.
- It is important to rehearse your report, think over how you will greet the audience, what you will say first, how you will finish the presentation. All comes with experience.
- Choose the right outfit, because. The speaker's clothing also plays a big role in the perception of his speech.
- Try to speak confidently, fluently and coherently.
- Try to enjoy the performance so you can be more relaxed and less anxious.
