Definition.
This is a hexagon whose bases are two equal square, and the side faces are equal rectangles
Side rib- this is common side two adjacent side faces
Prism Height- this is a cut perpendicular to bases prisms
Prism Diagonal- a segment connecting two vertices of the bases that do not belong to the same face
Diagonal plane is the plane that passes through the diagonal of the prism and its side ribs
Diagonal section - the boundaries of the intersection of the prism and the diagonal plane. Diagonal section of the correct quadrangular prism is a rectangle
Perpendicular section (orthogonal section)- this is the intersection of a prism and a plane drawn perpendicular to its side edges
Elements of a regular quadrangular prism
The figure shows two regular quadrangular prisms, which are marked with the corresponding letters:
- Bases ABCD and A 1 B 1 C 1 D 1 are equal and parallel to each other
- Side faces AA 1 D 1 D, AA 1 B 1 B, BB 1 C 1 C and CC 1 D 1 D, each of which is a rectangle
- Side surface- the sum of the areas of all side faces of the prism
- Total surface - the sum of the areas of all bases and side faces (the sum of the area of the side surface and bases)
- Side ribs AA 1 , BB 1 , CC 1 and DD 1 .
- Diagonal B 1 D
- Base diagonal BD
- Diagonal section BB 1 D 1 D
- Perpendicular section A 2 B 2 C 2 D 2 .
Properties of a regular quadrangular prism
- The bases are two equal squares
- The bases are parallel to each other
- The sides are rectangles.
- Side faces are equal to each other
- Side faces are perpendicular to the bases
- Lateral ribs are parallel to each other and equal
- Perpendicular section perpendicular to all side ribs and parallel to the bases
- corners perpendicular section- straight
- The diagonal section of a regular quadrangular prism is a rectangle
- Perpendicular (orthogonal section) parallel to the bases
Formulas for a regular quadrangular prism
Instructions for solving problems
When solving problems on the topic " regular quadrangular prism" implies that:Correct prism- a prism at the base of which lies regular polygon, and the side edges are perpendicular to the base planes. That is, a regular quadrangular prism contains at its base square. (see above the properties of a regular quadrangular prism) Note. This is part of the lesson with tasks in geometry (section solid geometry - prism). Here are the tasks that cause difficulties in solving. If you need to solve a problem in geometry, which is not here - write about it in the forum. To indicate the action of extracting square root symbol is used in problem solving√ .
A task.
In a regular quadrangular prism, the base area is 144 cm 2 and the height is 14 cm. Find the diagonal of the prism and the area full surface.Solution.
A regular quadrilateral is a square.
Accordingly, the side of the base will be equal to
Where does the diagonal of the base come from? rectangular prism will be equal to
√(12 2 + 12 2 ) = √288 = 12√2
Diagonal right prism forms with the diagonal of the base and the height of the prism right triangle. Accordingly, according to the Pythagorean theorem, the diagonal of a given regular quadrangular prism will be equal to:
√((12√2) 2 + 14 2 ) = 22 cm
Answer: 22 cm
A task
Find the total surface area of a regular quadrangular prism if its diagonal is 5 cm and the diagonal of the side face is 4 cm.Solution.
Since the base of a regular quadrangular prism is a square, then the side of the base (denoted as a) is found by the Pythagorean theorem:
A 2 + a 2 = 5 2
2a 2 = 25
a = √12.5
The height of the side face (denoted as h) will then be equal to:
H 2 + 12.5 \u003d 4 2
h 2 + 12.5 = 16
h 2 \u003d 3.5
h = √3.5
The total surface area will be equal to the sum of the lateral surface area and twice the base area
S = 2a 2 + 4ah
S = 25 + 4√12.5 * √3.5
S = 25 + 4√43.75
S = 25 + 4√(175/4)
S = 25 + 4√(7*25/4)
S \u003d 25 + 10√7 ≈ 51.46 cm 2.
Answer: 25 + 10√7 ≈ 51.46 cm 2.
It should be noted that combinatorics is an independent branch higher mathematics(and not part of terver) and weighty textbooks have been written in this discipline, the content of which, at times, is no easier than abstract algebra. However, a small fraction will suffice for us. theoretical knowledge, and in this article I will try to accessible form to analyze the basics of the topic with typical combinatorial problems. And many of you will help me ;-)
What are we going to do? AT narrow sense combinatorics is the counting of different combinations that can be made from a set discrete objects. Objects are understood as any isolated objects or living beings - people, animals, mushrooms, plants, insects, etc. At the same time, combinatorics does not care at all that the set consists of a plate of semolina, a soldering iron and a marsh frog. It is fundamentally important that these objects are enumerable - there are three of them. (discreteness) and it is essential that none of them are alike.
With a lot sorted out, now about the combinations. The most common types of combinations are permutations of objects, their selection from a set (combination) and distribution (placement). Let's see how this happens right now:
Permutations, combinations and placements without repetition
Do not be afraid of obscure terms, especially since some of them are really not very successful. Let's start with the tail of the title - what does " without repetition"? This means that in this paragraph sets will be considered, which consist of various objects. For example, ... no, I won’t offer porridge with a soldering iron and a frog, something tastier is better =) Imagine that an apple, a pear and a banana materialized on the table in front of you (if there are any, the situation can be simulated and real). We lay out the fruits from left to right in the following order:
apple / pear / banana
Question one: In how many ways can they be rearranged?
One combination has already been written above and there are no problems with the rest:
apple / banana / pear
pear / apple / banana
pear / banana / apple
banana / apple / pear
banana / pear / apple
Total: 6 combinations or 6 permutations.
Well, it was not difficult to list all possible cases here, but what if there are more items? Already with four different fruits, the number of combinations will increase significantly!
Please open reference material (Manual is easy to print) and in paragraph number 2, find the formula for the number of permutations.
No torment - 3 objects can be rearranged in ways.
Question two: In how many ways can you choose a) one fruit, b) two fruits, c) three fruits, d) at least one fruit?
Why choose? So they worked up an appetite in the previous paragraph - in order to eat! =)
a) One fruit can be chosen, obviously, in three ways - take either an apple, or a pear, or a banana. The formal count is based on formula for the number of combinations:
Recording in this case should be understood as follows: “in how many ways can you choose 1 fruit out of three?”
b) We list all possible combinations of two fruits:
apple and pear;
apple and banana;
pear and banana.
The number of combinations is easy to check using the same formula:
The entry is understood similarly: “in how many ways can you choose 2 fruits out of three?”.
c) And finally, three fruits can be chosen the only way:
By the way, the formula for the number of combinations also makes sense for an empty sample:
In this way, you can choose not a single fruit - in fact, take nothing and that's it.
d) In how many ways can you take at least one fruit? The “at least one” condition implies that we are satisfied with 1 fruit (any) or any 2 fruits or all 3 fruits:
ways you can choose at least one fruit.
Readers who have carefully studied introductory lesson on probability theory already figured something out. But about the meaning of the plus sign later.
For an answer to next question I need two volunteers ... ... Well, since no one wants, then I will call to the board =)
Question three: In how many ways can one fruit be distributed to Dasha and Natasha?
In order to distribute two fruits, you must first select them. According to paragraph "be" of the previous question, this can be done in ways, I will rewrite them again:
apple and pear;
apple and banana;
pear and banana.
But now there will be twice as many combinations. Consider, for example, the first pair of fruits:
you can treat Dasha with an apple, and Natasha with a pear;
or vice versa - Dasha will get the pear, and Natasha will get the apple.
And such a permutation is possible for every pair of fruits.
Consider the same student group who went to the dance. In how many ways can a boy and a girl be paired?
Ways you can choose 1 young man;
ways you can choose 1 girl.
So one young man and one girl can be chosen: 
ways.
When 1 object is selected from each set, then the following principle of counting combinations is valid: “ each an object from one set can form a pair with every object of another set.
That is, Oleg can invite any of the 13 girls to dance, Evgeny - also any of the thirteen, and other young people have a similar choice. Total: possible pairs.
It should be noted that in this example the "history" of pair formation does not matter; however, if initiative is taken into account, then the number of combinations must be doubled, since each of the 13 girls can also invite any boy to dance. It all depends on the conditions of a particular task!
A similar principle is valid for more complex combinations, for example: in how many ways can two young men be chosen and two girls to participate in a KVN skit?
Union And hints unambiguously that the combinations must be multiplied:
Possible groups of artists.
In other words, each pair of boys (45 unique pairs) can compete with any a couple of girls (78 unique couples). And if we consider the distribution of roles between the participants, then there will be even more combinations. ... I really want to, but still I will refrain from continuing so as not to instill in you an aversion to student life =).
The rule for multiplying combinations also applies to large quantity multipliers:
Task 8
How many three-digit numbers are there that are divisible by 5?
Solution: for clarity, we denote given number three stars: ***
AT hundreds place you can write any of the numbers (1, 2, 3, 4, 5, 6, 7, 8 or 9). Zero is not good, because in this case the number ceases to be three-digit.
But in tens place(“in the middle”) you can choose any of the 10 digits: .
By condition, the number must be divisible by 5. The number is divisible by 5 if it ends in 5 or 0. Thus, in the least significant digit, we are satisfied with 2 digits.
Total, there is: three-digit numbers that are divisible by 5.
At the same time, the work is deciphered as follows: “9 ways you can choose a number in hundreds place and 10 ways to select a number in tens place and 2 ways in unit digit»
Or even simpler: each from 9 digits to hundreds place combined with each of 10 digits tens place and with each of two digits units digit».
Answer: 180
And now…
Yes, I almost forgot about the promised commentary to problem No. 5, in which Borya, Dima and Volodya can be dealt one card each in different ways. Multiplication here has the same meaning: in ways you can extract 3 cards from the deck And in each sample to rearrange them ways.
Now the task for independent decision... now I'll come up with something more interesting, ... let it be about the same Russian version of blackjack:
Task 9
How many winning combinations of 2 cards are there in a "point" game?
For those who don't know: wins combination 10 + ACE (11 points) = 21 points and, let's consider the winning combination of two aces.
(the order of the cards in any pair does not matter)
Quick Solution and the answer at the end of the lesson.
By the way, it is not necessary to consider an example primitive. Blackjack is almost the only game for which there is a mathematically justified algorithm that allows you to beat the casino. Those who wish can easily find a lot of information about the optimal strategy and tactics. True, such masters quickly fall into the black list of all establishments =)
It's time to consolidate the material covered with a couple of solid tasks:
Task 10
Vasya has 4 cats at home.
a) In how many ways can the cats be seated in the corners of the room?
b) In how many ways can cats be allowed to roam?
c) in how many ways can Vasya pick up two cats (one on the left, the other on the right)?
We decide: first, it should again be noted that the problem is about different objects (even if the cats are identical twins). This is very important condition!
a) Silence of cats. This execution is subject to all cats at once
+ their location is important, so there are permutations here:
ways you can seat cats in the corners of the room.
I repeat that in permutations, only the number of different objects and their mutual arrangement. Depending on his mood, Vasya can seat the animals in a semicircle on the sofa, in a row on the windowsill, etc. - there will be 24 permutations in all cases. For convenience, those who wish can imagine that the cats are multi-colored (for example, white, black, red and striped) and list all possible combinations.
b) In how many ways can cats be allowed to roam?
It is assumed that cats go for a walk only through the door, while the question implies indifference about the number of animals - 1, 2, 3 or all 4 cats can go for a walk.
We consider all possible combinations:
Ways you can let go for a walk one cat (any of the four); 
ways you can let two cats go for a walk (list the options yourself);
ways you can let three cats go for a walk (one of the four sits at home);
way you can release all the cats.
You probably guessed that the obtained values should be summed up:
ways to let cats go for a walk.
For enthusiasts, I offer a complicated version of the problem - when any cat in any sample can randomly go outside, both through the door and through the window of the 10th floor. There will be more combinations!
c) In how many ways can Vasya pick up two cats?
The situation involves not only the choice of 2 animals, but also their placement on the hands:
ways you can pick up 2 cats.
The second solution: in ways you can choose two cats and ways to plant every a couple in hand: 
Answer: a) 24, b) 15, c) 12
Well, to clear my conscience, something more specific on the multiplication of combinations .... Let Vasya have 5 extra cats =) How many ways can you let 2 cats go for a walk and 1 cat?
That is, with each a couple of cats can be released every cat.
Another button accordion for an independent solution:
Task 11
3 passengers got into the elevator of a 12-storey building. Everyone, independently of the others, can exit on any (starting from the 2nd) floor with the same probability. In how many ways:
1) Passengers can get off at the same floor (exit order doesn't matter);
2) two people can get off on one floor and a third on another;
3) people can get off at different floors;
4) Can passengers exit the elevator?
And here they often ask again, I clarify: if 2 or 3 people go out on the same floor, then the order of exit does not matter. THINK, use formulas and rules for addition/multiplication combinations. In case of difficulty, it is useful for passengers to give names and reason in what combinations they can get out of the elevator. There is no need to be upset if something does not work out, for example, point number 2 is quite insidious.
Complete Solution with detailed comments at the end of the lesson.
The final paragraph is devoted to combinations that are also quite common - in my opinion subjective assessment, about 20-30% combinatorial problems:
Permutations, combinations and placements with repetitions
Listed species combinations are outlined in paragraph number 5 reference material Basic formulas of combinatorics, however, some of them may not be very clear on first reading. In this case, it is advisable to first familiarize yourself with practical examples, and only then comprehend the general formulation. Go:
Permutations with repetitions
In permutations with repetitions, as in "ordinary" permutations, the whole set of objects at once, but there is one thing: in this set, one or more elements (objects) are repeated. Meet the next standard:
Task 12
How many different letter combinations can be obtained by rearranging cards with the following letters: K, O, L, O, K, O, L, L, H, I, K?
Solution: in the event that all letters were different, then a trivial formula should be applied, however, it is quite clear that for the proposed set of cards, some manipulations will work "idle", so, for example, if you swap any two cards with the letters "K in any word, it will be the same word. Moreover, physically the cards can be very different: one can be round with a printed letter “K”, the other is square with a drawn letter “K”. But according to the meaning of the problem, even such cards considered the same, since the condition asks about letter combinations.
Everything is extremely simple - in total: 11 cards, including the letter:
K - repeated 3 times;
O - repeated 3 times;
L - repeated 2 times;
b - repeated 1 time;
H - repeated 1 time;
And - repeats 1 time.
Check: 3 + 3 + 2 + 1 + 1 + 1 = 11, which is what we wanted to check.
According to the formula number of permutations with repetitions:
various letter combinations can be obtained. More than half a million!
To quickly calculate a large factorial value, it is convenient to use standard function Excel: we score in any cell =FACT(11) and click Enter.
In practice, it is perfectly acceptable not to write general formula and, in addition, omit the unit factorials: 
But preliminary comments about repeated letters are required!
Answer: 554400
Another typical example of permutations with repetitions occurs in the placement problem chess pieces which can be found in stock ready-made solutions in the corresponding pdf. And for an independent solution, I came up with a less template task:
Task 13
Alexey goes in for sports, and 4 days a week - athletics, 2 days - strength exercises and 1 day rest. In how many ways can he schedule his weekly classes?
The formula doesn't work here because it takes into account overlapping permutations (for example, when strength exercises on Wednesday are swapped with strength exercises on Thursday). And again - in fact, the same 2 strength training sessions can be very different from each other, but in the context of the task (in terms of the schedule), they are considered the same elements.
Two-line solution and answer at the end of the lesson.
Combinations with repetitions
Feature This type of combination is that the sample is drawn from several groups, each of which consists of the same objects.
Everyone worked hard today, so it's time to refresh yourself:
Task 14
The student cafeteria sells sausages in dough, cheesecakes and donuts. In how many ways can five cakes be purchased?
Solution: immediately pay attention to the typical criterion for combinations with repetitions - according to the condition, not a set of objects as such, but different kinds objects; it is assumed that there are at least five hot dogs, 5 cheesecakes and 5 donuts on sale. The pies in each group, of course, are different - because absolutely identical donuts can only be modeled on a computer =) However physical characteristics pies are not essential in the sense of the problem, and hot dogs / cheesecakes / donuts in their groups are considered the same.
What can be in the sample? First of all, it should be noted that the sample will necessarily contain identical pies(since we choose 5 pieces, and there are 3 types to choose from). Options here for every taste: 5 hot dogs, 5 cheesecakes, 5 donuts, 3 hot dogs + 2 cheesecakes, 1 hot dog + 2 + cheesecakes + 2 donuts, etc.
As with "regular" combinations, the order of selection and placement of pies in the sample does not matter - they just chose 5 pieces and that's it.
We use the formula 
number of combinations with repetitions: 
way you can buy 5 pies.
Answer: 21
What conclusion can be drawn from many combinatorial problems?
Sometimes, the most difficult thing is to understand the condition.
A similar example for a do-it-yourself solution:
Task 15
There is enough in the wallet a large number of 1-, 2-, 5- and 10-ruble coins. In how many ways can three coins be taken out of the wallet?
For self-control, answer a couple simple questions:
1) Can all coins in the sample be different?
2) Name the "cheapest" and the most "expensive" combination of coins.
Solution and answers at the end of the lesson.
From my personal experience, I can say that combinations with repetitions are the rarest guest in practice, which cannot be said about following form combinations:
Placements with repetitions
From a set consisting of elements, elements are selected, and the order of the elements in each sample is important. And everything would be fine, but a rather unexpected joke is that we can choose any object of the original set as many times as we like. Figuratively speaking, from "the multitude will not decrease."
When does it happen? Typical example is a combination lock with several disks, but due to the development of technology, it is more relevant to consider its digital descendant:
Task 16
How many 4-digit pin codes are there?
Solution: in fact, to solve the problem, it is enough to know the rules of combinatorics: you can choose the first digit of the pin code in ways and ways - the second digit of the pin code and in as many ways - a third and as many - the fourth. Thus, according to the rule of multiplication of combinations, a four-digit pin code can be composed: in ways.
And now with the formula. By condition, we are offered a set of numbers, from which numbers are selected and placed in a certain order, while the numbers in the sample can be repeated (i.e. any digit of the original set can be used an arbitrary number of times). According to the formula for the number of placements with repetitions: 
Answer: 10000
What comes to mind here ... ... if the ATM "eats" the card after the third failed attempt entering a PIN code, then the chances of picking it up at random are very illusory.
And who said that there is no practical sense in combinatorics? A cognitive task for all readers of the site:
Problem 17
According to state standard, a car license plate consists of 3 numbers and 3 letters. In this case, a number with three zeros is not allowed, and the letters are selected from the set A, B, E, K, M, H, O, R, C, T, U, X (only those Cyrillic letters are used, the spelling of which matches the Latin letters).
How many different license plates can be composed for a region?
Not so, by the way, and a lot. AT major regions this number is not enough, and therefore for them there are several codes for the inscription RUS.
Solution and answer at the end of the lesson. Don't forget to use the rules of combinatorics ;-) …I wanted to brag about being exclusive, but it turned out to be not exclusive =) I looked at Wikipedia - there are calculations there, however, without comments. Although in educational purposes, probably, few people solved it.
Our an exciting activity came to an end, and in the end I want to say that you did not waste your time - for the reason that the combinatorics formulas find one more vital practical use: they meet in various tasks on probability theory,
and in tasks on the classical definition of probability- especially often
Thank you all for Active participation and see you soon!
Solutions and answers:
Task 2: Solution: find the number of all possible permutations of 4 cards:
When a card with a zero is in 1st place, the number becomes three-digit, so these combinations should be excluded. Let zero be in the 1st place, then the remaining 3 digits in the least significant digits can be rearranged in ways.
Note
: because there are few cards, it is easy to list all such options here:
0579
0597
0759
0795
0957
0975
Thus, from the proposed set, you can make:
24 - 6 = 18 four-digit numbers
Answer
: 18
Task 4: Solution: 3 cards can be selected from 36 ways.
Answer
: 7140
Task 6: Solution: 
ways.
Another solution
: ways you can select two people from the group and and
2) The “cheapest” set contains 3 ruble coins, and the most “expensive” set contains 3 ten-ruble coins.
Task 17: Solution: 
ways you can make a digital combination of a license plate, while one of them (000) should be excluded:.
ways you can make a letter combination of a car number.
According to the rule of multiplication of combinations, everything can be composed:
car numbers
(each digital combination combined with each letter combination).
Answer
: 1726272
Combinatorial tasks
1 . Katya, Masha and Ira are playing with a ball. Each of them must throw the ball once in the direction of each friend. How many times should each of the girls throw the ball? How many times will the ball be tossed? Determine how many times the ball will be tossed if the game is attended by: four children; five children.
2 . Given three facades and two roofs that have the same shape but are painted in different colors: the facades are yellow, blue and red, and the roofs are blue and red. What houses can be built? How many combinations are there?
3 . Given three identical facades of the house: blue, yellow and red - and three roofs: blue, yellow and red. What houses can be built? How many combinations are there?
4 . Flags may be circle, square, triangle, or star, and may be colored green or red. How many different flags can there be?
5. In the school canteen for lunch, meat, meatballs and fish were prepared as second courses. For dessert - ice cream, fruit and pie. You can choose one main course and one dessert course. How many exist various options lunch?
6. In the school canteen for lunch, soup with meat and vegetarian soup were prepared as first courses, meat, meatballs and fish for the second, ice cream, fruit and a pie for sweets. How many different options for a three-course meal are there?
7. In how many ways can three students be seated in a row on chairs? List all possible cases.
8 . In how many ways can four (five) people line up?
9 . FROM different parties Three paths rise up the hill and converge at the top. Make up many routes that you can take up and down the hill. Solve the same problem if you have to go up and down different paths.
10 . Three roads lead from Akulovo to Rybnitsa, and four roads lead from Rybnitsa to Kitovo. How many ways are there to travel from Akulovo to Kitovo via Rybnitsa?
11 . A syllable is called open if it starts with a consonant and ends with a vowel. How many open two-letter syllables can be written using the letters "a", "b", "c", "d", "e", "i", "o"? Write out these syllables.
12. How many different blouse and skirt suits can be made if there are 4 blouses and 4 skirts?
13. When Petya goes to school, he sometimes meets one or more of his friends: Vasya, Lenya, Tolya. List all the possible cases that this may be.
14 . Write down all possible two-digit numbers using the numbers 7 and 4.
15 . Misha planned to buy: a pencil, a ruler, a notebook and a notebook. Today he only bought two different subject. What could Misha buy, assuming that the store had all the educational supplies he needed?
16 . The four people shook hands. How many handshakes were there?
17 . How many two-digit numbers are there without the digit 0?
18 . Write down all possible three-digit numbers that can be formed from the numbers 1 and 2.
19 . Write down all possible even three-digit numbers made up of the numbers 1 and 2.
20 . Write down all possible two-digit numbers that use the numbers 2, 8 and 5.
21 . How many different two-digit numbers are there, all of whose digits are odd?
22 . What three-digit numbers can be written using the numbers 3, 7 and 1, provided that the number must not contain the same digits? How many such numbers?
23 . How three-digit numbers can be made up of the numbers 1, 2, 4, 6, if no number is used more than once? How many of these numbers will be even? How many odd ones?
24 . The car has five seats. In how many ways can five people get into this car if only two of them can take the driver's seat?
25. There are 5 single desks in the class. In how many ways can two (three) newly arrived schoolchildren be seated on them?
26 . Remember I. Krylov's fable "Quartet":
The naughty Monkey, the Donkey, the Goat and the clubfoot Mishka started to play the Quartet. They hit the bows, they tear, but there is no sense. “Stop, brothers, stop! - shouts Monkey. - Wait! How does the music go? You don't sit like that." How many different ways can these musicians try to sit down? Can it improve the quality of their game?
27 . Boys and girls are seated in a row in consecutive places, with the boys sitting in odd-numbered places, and girls in even-numbered places. In how many ways can this be done if:
a) 3 boys and 3 girls are seated in 6 places;
b) 5 boys and 5 girls are seated in 10 seats?
28 . Two checkers must be placed on an empty checkerboard - black and white. How many different positions can they occupy on the board?
29. Let the car number be composed of two letters followed by two numbers, for example AB-53. How many different numbers can be made using 5 letters and 6 numbers?
30 . The vehicle number consists of three letters and four numbers. How many different license plates(three letters are taken from 29 letters of the Russian alphabet)?
31 . Suppose you had to go to the library, the savings bank, the post office and give shoes for repair. In order to choose the shortest route, it is necessary to consider all possible options. How many ways are there if the library, the savings bank, the post office, and the shoe shop are far apart?
32. Suppose you had to go to the library, the savings bank, the post office and give shoes for repair. In order to choose the shortest route, it is necessary to consider all possible options. How many reasonable paths are there if the library and the post office are close by, but far removed from the savings bank and the shoe shop, which are far apart?
33. There was a lively discussion of the four magazines among the passengers in the carriage. It turned out that everyone subscribes to two magazines, and each of the possible combinations of two magazines is subscribed by one person. How many people were in this group?
34 . There are five dice that differ from each other only in color: 2 red, 1 white and 2 black. There are two boxes A and B, where A holds 2 cubes and B holds 3. In how many different ways can these cubes be placed in boxes A and B?
35. In order to bring rejuvenating apples to the Tsar Father, Ivan Tsarevich must find the only true path to the magic garden. I met Ivan Tsarevich at the fork in the three roads of the old raven and this is what advice I heard from him:
1) go now along the right path;
2) on next fork do not choose the right path;
3) at the third fork, do not take the left path.
A dove flying past whispered to Ivan Tsarevich that only one advice from the raven was correct and that one must definitely go along the paths in different directions. Our hero completed the task and ended up in a magical garden. What route did he take?
I offer readers of "Habrahabr" a translation of the publication "100 Prisoners Escape Puzzle", which I found on the website of DataGenetics. Please send all errors in this article in private messages.
According to the condition of the problem, there are 100 prisoners in the prison, each of which has a personal number from 1 to 100. The jailer decides to give the prisoners a chance for release and offers to pass the test he invented. If all the prisoners succeed, then they are free, if at least one fails, they all die.
A task
The jailer goes to secret room and prepares 100 boxes with lids. On each box, he marks numbers from 1 to 100. Then he brings 100 paper tablets, according to the number of prisoners, and numbers these tablets from 1 to 100. After that, he shuffles 100 tablets and places one tablet in each box, closing the lid . The prisoners do not see how the jailer performs all these actions.The competition begins, the jailer takes each prisoner one by one to the room with the boxes and tells the prisoners that they must find a box that will contain a plate with the number of the prisoner. The prisoners are trying to find the plate with their number by opening the boxes. Each is allowed to open up to 50 boxes; if each of the prisoners finds his number, then the prisoners will be released, if at least one of them does not find his number in 50 attempts, then all the prisoners will die.
In order for prisoners to be released, ALL prisoners must pass the test successfully.
So what is the chance that the prisoners will be pardoned?
- After the prisoner opens the box and checks the plate, it is placed back in the box and the lid is closed again;
- Places of the plates cannot be changed;
- Prisoners cannot leave clues to each other or interact with each other in any way once the trial has begun;
- Prisoners are allowed to discuss strategy before the trial begins.
What is the optimal strategy for prisoners?
Additional question:
If a friend of the prisoners (not a participant in the test) will be able to enter the secret room before the start of the test, examine all the tablets in all boxes and (optionally, but not required) swap two tablets from two boxes (in this case, the comrade will not have the opportunity to both inform the prisoners about the result of his actions), then what strategy should he take to increase the chances of the prisoners to escape?
Solution improbable?
At first glance, this task seems almost hopeless. It seems that the chance for each of the prisoners to find their tablet is microscopically small. In addition, prisoners cannot exchange information with each other during the trial.The odds of one prisoner are 50:50. There are 100 boxes in total and he can open up to 50 boxes looking for his sign. If he opens the boxes at random and opens half of all the boxes, he will find his tablet in the open half of the boxes, or his tablet will remain in the closed 50 boxes. His chances of success are ½.
Let's take two prisoners. If both choose boxes at random, the chances for each of them will be ½, and for two ½x½=¼.
(for two prisoners, success will be in one case out of four).
For three prisoners, the odds are ½ × ½ × ½ = ⅛.
For 100 prisoners, the odds are: ½ × ½ × … ½ × ½ (multiply 100 times).
This equals Pr ≈ 0.0000000000000000000000000000008
So it's a very small chance. In this scenario, most likely, all the prisoners will be dead.
Incredible answer
If each prisoner opens the boxes at random, they are unlikely to pass the test. There is a strategy where prisoners can expect to be successful more than 30% of the time. This is a stunningly incredible result (if you haven't heard of this math problem previously).More than 30% for all 100 prisoners! Yes, this is even more than the chances for two prisoners, provided that they open the boxes at random. But how is this possible?
It is clear that one for each prisoner, the chances cannot be higher than 50% (after all, there is no way for communication between prisoners). But do not forget that the information is stored in the location of the plates inside the boxes. No one shuffles the tablets between visits to the room by individual prisoners, so we can use that information.
Solution
First, I'll tell you the solution, then I'll explain why it works.The strategy is extremely easy. The first of the prisoners opens the box with the number that is written on his clothes. For example, prisoner number 78 opens the box with the number 78. If he finds his number on the plate inside the box, that's great! If not, he looks at the number on the plate in "his" box and then opens the next box with that number. Having opened the second box, he looks at the number of the tablet inside this box and opens the third box with this number. Then we simply transfer this strategy to the remaining boxes. For clarity, look at the picture:
Eventually, the prisoner will either find his number or reach the 50 box limit. At first glance, this seems pointless compared to simply choosing a box at random (and for one individual prisoner it does), but since all 100 prisoners will be using the same set of boxes, it makes sense.
The beauty of this math problem- not only to know the result, but also to understand why this strategy works.
So why does the strategy work?
Each box contains one plate - and this plate is unique. This means that the plate is in a box with the same number, or it points to a different box. Since all plates are unique, there is only one plate for each box pointing to it (and only one way to get to that box).
If you think about it, the boxes form a closed circular chain. One box can be part of only one chain, since inside the box there is only one pointer to the next one and, accordingly, in the previous box there is only one pointer to this box (programmers can see the analogy with linked lists).
If the box does not point to itself (the box number is equal to the plate number in it), then it will be in the chain. Some chains may consist of two boxes, some are longer.
Since all prisoners start with a box with the same number on their clothes, they are, by definition, placed on the chain that contains their nameplate (there is only one nameplate that points to this box).
Exploring the boxes along this chain in a circle, they are guaranteed to eventually find their sign.
The only question remains whether they will find their tablet in 50 moves.
Chain length
In order for all prisoners to pass the test, the maximum chain length must be less than 50 boxes. If the chain is longer than 50 boxes, prisoners with numbers from those chains will fail the test - and all prisoners will be dead.If the maximum length of the longest chain is less than 50 boxes, then all prisoners will pass the test!
Think about it for a second. It turns out that there can only be one chain that is longer than 50 boxes in any layout of the plates (we have only 100 boxes, so if one chain is longer than 50, then the rest will be shorter than 50 in total).
Long chain hand odds
Once you've convinced yourself that the maximum chain length must be less than or equal to 50 to succeed, and there can only be one long chain in any set, we can calculate the probability of passing the challenge:Some more math
So what do we need to figure out the probability of a long chain?For a chain of length l, the probability that the boxes will be outside this chain is:
There is (l-1) in this collection of numbers! ways to arrange the signs.
The remaining signs can be located (100-l)! ways (do not forget that the length of the chain does not exceed 50).
Given this, the number of permutations that contain the chain exact length l: (>50)
It turns out that there are 100(!) ways to arrange the plates, so that the probability of existence of a chain of length l is equal to 1/l. By the way, this result does not depend on the number of boxes.
As we already know, there can only be one case in which there is a chain with a length > 50, so the probability of success is calculated by this formula:
Result
31.18% - the probability that the size of the longest chain will be less than 50 and each of the prisoners will be able to find their tablet, given the limit of 50 attempts.The probability that all prisoners will find their plates and pass the test is 31.18%
Below is a graph showing the probabilities (on the y-axis) for all chains of length l (on the x-axis). Red means all "failures" (given curve here is just a 1/l plot). Green color means "success" (the calculation is a little more complicated for this part of the graph, since there are several ways to determine maximum length <50). Общая вероятность складывается из зеленых столбцов в 31.18% шанс на спасение.
Harmonic number (this part of the article is for geeks)
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n consecutive numbers of the natural series.
Let's calculate the limit if instead of 100a boxes we have an arbitrary large number of boxes (let's assume that we have 2n boxes in total).
The Euler-Mascheroni constant is a constant defined as the limit of the difference between the partial sum of a harmonic series and the natural logarithm of a number.
As the number of prisoners increases, if the overseer allows prisoners to open half of all the boxes, then the chance of salvation tends to 30.685%
(If you made a decision in which the prisoners randomly guess the boxes, then as the number of prisoners increases, the probability of being saved tends to zero!)
Additional question
Anyone else remember the extra question? What can our helpful comrade do to increase our chances of survival?Now we already know the solution, so the strategy here is simple: he must examine all the signs and find the longest chain of boxes. If the longest chain is less than 50, then he does not need to change the tablets at all, or change them so that the longest chain does not become longer than 50. However, if he finds a chain longer than 50 boxes, all he has to do is swap the contents of two boxes from that chain to break that chain into two shorter chains.
As a result of this strategy, there will be no long chains and all prisoners are guaranteed to find their sign and salvation. So, by swapping two signs, we reduce the probability of salvation to 100%!
2017-2018 Training work in mathematics Grade 11
Option 2 (basic)
The answer to each task is a final decimal fraction, an integer or a sequence of digits. Write down the answers to the tasks in the answer field in the text of the work, and then transfer them to the answer form No. 1 to the right of the number of the corresponding task. If the answer is a sequence of numbers, then write down this sequence in the answer sheet No. 1without spaces, commas and other additional characters. Write each number, minus sign and comma in a separate box. Units of measurement are not required.
1
Answer: _________________.
2 . Find the value of the expression:
Answer: _________________.
3 . At school, girls make up 51% of all students. How many girls are in this school if there are 8 more girls than boys?
Answer: _________________.
4 . Harmonic mean of three numbersa , b andWith, is calculated by the formula Find the harmonic mean of numbers
Answer: _________________.
5. Calculate:
Answer: _________________.
6 . In the men's dormitory of the institute, no more than three people can be accommodated in each room. What is the smallest number of rooms needed to accommodate 79 out-of-town students?
Answer: _________________.
7 .Find the root of the equation
Answer: _________________.
8 . The apartment consists of two rooms, a kitchen, a corridor and a bathroom (see drawing). The first room has 4 m by 4 m, the second - 4 m by 3.5 m, the kitchen has dimensions of 4 m by 3.5 m, the bathroom - 1.5 m by 2 m. Find the area of the corridor. Give your answer in square meters.
Answer: _________________.
9 . Establish a correspondence between the quantities and their possible values: for each element of the first column, select the corresponding element from the second column.
VALUE VALUES
A) the volume of the chest of drawers 1) 0.75 l
B) the volume of water in the Caspian Sea 2) 78200 km 3
C) the volume of the package of ryazhenka 3) 96 l
D) the volume of the railway car 4) 90 m 3
In the table, under each letter corresponding to the value, indicate the number of its possible value.
Answer:
Answer: _________________.10 . At the Russian Language Olympiad, participants are seated in three classrooms. In the first two, 130 people each, the rest are taken to a reserve auditorium in another building. When counting, it turned out that there were 400 participants in total. Find the probability that a randomly selected participant wrote the Olympiad in the spare room.
Answer: _________________.
11 . The figure shows a graph of atmospheric pressure values in a certain city for three days. Days of the week and time are indicated horizontally, atmospheric pressure values in millimeters of mercury are indicated vertically. Find the value of atmospheric pressure on Wednesday at 12 o'clock. Give your answer in millimeters of mercury.
Answer: ____________.
Answer: _________________.
13. A regular hexagonal pyramid with edge 1 was glued to a regular hexagonal prism with edge 1 so that the faces of the bases coincided. How many faces does the resulting polyhedron have (invisible edges are not shown in the figure)?
Answer: _________________.
14. The figure shows a graph of the function pointsA, B, C, DandEset on the axisX four intervals. Using the graph, match each interval with the characteristic of a function or its derivative.
INTERVALS OF THE CHARACTERISTIC OF A FUNCTION OR DERIVATIVE
A) (A; B) 1) the function changes sign from “-” to “+”
B) (C; C) 2) the derivative changes sign from "-" to "+"
B) (C;D) 3) the derivative changes sign from "+" to "-"
G) (D; E) 4) the function is positive and increasing
In the table below each letter, indicate the corresponding number.
15 . On a circle with a centerO points are markedBUT andAT so that the length of the smaller arcAB is 3. Find the length of the larger arc.
Answer: _________________.
16 . Given two boxes that have the shape of a regular quadrangular prism. The first box is four and a half times lower than the second, and the second is three times narrower than the first. How many times greater is the volume of the first box than the volume of the second?
Answer: _________________.
17. Each of the four inequalities in the left column corresponds to one of the solutions in the right column. Establish a correspondence between inequalities and their solutions.
INEQUALITY OF SOLUTIONS
BUT)
B)
AT)
G)
Write in the table given in the answer under each letter the corresponding number of the decision.
Answer:
18 . At the Winter Olympics, the Russian team won more medals than the Canadian team, the Canadian team - more than the German team, and the Norwegian team - less than the Canadian team.Select the statements that are true under the given conditions.
1) Of the named teams, the Canadian team came second in the number of medals.
2) Among the named teams there are three that won an equal number of medals.
3) The German team won more medals than the Russian team.
4) The Russian team won more medals than each of the other three teams.
In your answer, indicate the numbers of correct statements in ascending order.
Answer: _________________.
19 . chetythree-digit numberBUT consists of numbers 3; four; eight; 9, afourthree-digit numberAT - from the numbers 6; 7; eight; 9. It is known thatAT = 2 BUT. Find a numberBUT. In your answer, indicate any one such number, except for the number 3489.
Answer: _________________.
20 . The rectangle is divided into four small rectangles by two straight cuts. The perimeters of three of them, starting from the top left and going clockwise, are 17, 15, and 18. Find the perimeter of the fourth rectangle.
1715
?
18
