Combinatorics - this, as the name itself suggests, is a branch of mathematics that studies various sets or combinations any objects (elements) - numbers, objects, letters in words, etc. A very interesting section.) But for one reason or another, difficult to understand. Why? Because it often contains terms and designations that are more difficult for visual perception. If the characters are 10, 2, 3/4 and even, or log 2 5 are visually clear to us, i.e. we can somehow “feel” them, then with designations like 15!,P 9 . . problems begin. In addition, in most textbooks this topic is presented rather dryly and difficult to understand. I hope this material will help solve these problems at least a little and you will like combinatorics.)
Each of us faces combinatorial problems every day. When we decide in the morning how to dress, we combine certain types of clothing. When we prepare a salad, we combine the ingredients. The result depends on what combination of products is chosen - tasty or tasteless. True, issues of taste are no longer dealt with by mathematics, but by cooking, but still.) When we play “words”, making small words from one long one, we combine letters. When we open a combination lock or dial a phone number, we combine the numbers.) The head teacher of the school draws up lesson schedules, combining subjects. Football teams at the World or European Championships are divided into groups, forming combinations. And so on.)
People solved combinatorial problems in ancient times (magic squares, chess), and the real heyday of combinatorics occurred in the 6th–7th centuries, during the widespread use of gambling (cards, dice), when players had to think through various moves and thus actually also solve combinatorial problems.) Along with combinatorics, another branch of mathematics arose at the same time - probability theory . These two sections are very close relatives and go hand in hand.) And when studying probability theory, we will more than once encounter combinatorics problems.
And we will begin the study of combinatorics with such a cornerstone concept as factorial .
What is factorial?
The word “factorial” is a beautiful word, but it frightens and confuses many. But in vain. In this lesson we will understand and work well with this simple concept.) This word comes from the Latin “factorialis”, which means “multiplying”. And for good reason: the calculation of any factorial is based on the ordinary multiplication.)) So, what is factorial.
Let's take some natural number n . Completely arbitrary: we want 2, we want 10, whatever, as long as it’s natural.) So, factorial of a natural number n is the product of all natural numbers from 1 to n inclusive. It is designated like this: n! That is,
In order not to describe this long work every time, we simply came up with a short notation. :) It reads a little unusually: “en factorial” (and not the other way around, “factorial en”, as it might seem).
That's all! For example,
Do you get the idea?)) Great! Then we consider examples:
Answers (in disarray): 30; 0.1; 144; 6; 720; 2; 5040.
Everything worked out? Wonderful! We already know how to calculate factorials and solve simple examples with them. Go ahead. :)
Properties of factorial
Let's consider the expression 0, which is not very clear from the point of view of determining the factorial. So in mathematics it was agreed that
Yes Yes! This is an interesting equation. Both from one and from zero the factorial is the same - one.)) For now, let’s take this equality as a dogma, but why this is exactly so will be clear a little later, with examples.))
The following two are very similar properties:
They can be proven in an elementary way. Directly in the meaning of factorial.)
These two formulas allow, firstly, to easily calculate the factorial of the current natural number through the factorial previous numbers. Or the next one through the current one.) Such formulas in mathematics are called recurrent.
Secondly, with the help of these formulas you can simplify and calculate some tricky expressions with factorials. Like these.
Calculate:
How will we proceed? Consistently multiply all natural numbers from 1 to 1999 and from 1 to 2000? You'll be stunned by this! But the properties of the example are solved literally in one line:
Or like this:
Or such a task. Simplify:
Again we work directly on the properties:
Why are factorials needed and where did they come from? Well, why are they needed? This is a philosophical question. In mathematics, nothing happens just for the sake of beauty.)) In fact, factorial has a great many applications. This is Newton’s binomial, and probability theory, and series, and Taylor’s formula, and even the famous numbere , which is an interesting infinite sum:
The more you askn , those larger number terms in the sum and the closer this sum will be to the numbere . And in limit when it becomes equal to exactly the numbere . :) But we will talk about this amazing number in the appropriate topic. And here we have factorials and combinatorics.)
Where did they come from? They came from combinatorics, from the study of sets of elements.) The simplest such set is rearrangement without repetition. Let's start with it. :)
Rearrangement without repetition
Let us have two various object. Or element. Absolutely any. Two apples (red and green), two candies (chocolate and caramel), two books, two numbers, two letters - anything. If only they were various.) Let's call themA AndB respectively.
What can you do with them? If these are candies, then, of course, you can eat them.)) We will tolerate them for now and eat them arrange in different order.
Each such location is called rearrangement without repetition. Why "no repetition"? Because all the elements involved in the permutation are different. For the sake of simplicity, we have decided this so far. Is there some more permutation with repetitions, where some elements may be the same. But such permutations are a little more complicated. More on them later.)
So, if two different elements are considered, then the following options are possible:
AB , B A .
There are only two options, i.e. two permutations. Not much.)
Now let's add one more element to our setC . In this case, there will be six permutations:
ABC , ACB , BAC , B.C.A. , CAB , C.B.A. .
We will construct permutations of four elements as follows. First, let's put the element firstA . At the same time, the remaining three elements can be rearranged, as we already know, six ways:
This means that the number of permutations with the first elementA equals 6.
But the same story will turn out if we put first any of these four elements. They have equal rights and each deserves to be in first place.) This means that the total number of permutations of four elements will be equal to . Here they are:
So, to summarize: permutation from n elements are called any ordered set of these nelements.
The word "ordered" is key here: each permutation differs only order of elements, and the elements themselves in the set remain the same.
It remains only to find out what the number of such permutations from any number of elements: we are not masochists to write out every time All various options and count them. :) For 4 elements we received 24 permutations - this is already quite a lot for visual perception. What if there are 10 elements? Or 100? It would be nice to construct a formula that, in one fell swoop, would count the number of all such permutations for any number of elements. And there is such a formula! Now we will derive it.) But first, let’s formulate one very important auxiliary rule in all combinatorics, called product rule .
Product rule: if included in the set n different options for choosing the first element and for each of them there is m different options for choosing the second element, then a total of n·m different pairs of these elements.
And now, let now there be a set ofn various elements
,
where, of course, . We need to count the number of all possible permutations of the elements of this set. We reason in exactly the same way.)) You can put any of these in first placen elements. It means that the number of ways to select the first element is n .
Now imagine that we have the first element selected (n ways, as we remember). How many unselected elements are left in the set? Right,n-1 . :) This means that the second element can only be selectedn-1 ways. Third -n-2 ways (since 2 elements are already selected). And so on, kth element can choosen-(k-1) ways, the penultimate one - in two ways, and the last element - in only one way, since all other elements are already selected in one way or another. :)
Well, now let's construct the formula.
So, the number of ways to select the first element from the set isn . On every of thesen ways according ton-1 way to choose the second one. This means that the total number of ways to select the 1st and 2nd elements, according to product rule, will be equaln(n-1) . Further, each of them, in turn, accounts forn-2 way to select the third element. Means, three element can already be selectedn(n-1)(n-2) ways. And so on:
4 elements - 
ways
k elements in ways,
n elements in ways.
Means, nelements can be selected (or in our case arranged) in ways.
The number of such methods is indicated as follows:P n . It reads: “pe from en.” From the French " P ermutation - rearrangement." Translated into Russian it means: "permutation from n elements".
Means,
Now let's look at the expression, standing on the right side of the formula. Doesn't remind you of anything? What if you rewrite it from right to left, like this?
Well, of course! Factorial, in person. :) Now you can briefly write down:
Means, number everyone possible permutations from n different elements are equal n! .
This is the main practical meaning of factorial.))
Now we can easily answer many questions related to combinations and permutations.)
In how many ways can 7 different books be placed on a shelf?
P 7 = 7! = 1 2·3·4·5·6·7 = 5040 ways.)
In how many ways can you make a schedule (for one day) from 6 different subjects?
P6 = 6! = 1 2·3·4·5·6 = 720 ways.
In how many ways can 12 people be arranged in a column?
No problem! P 12 = 12! = 1 2·3·...·12 = 479001600 ways. :)
Great, right?
There is one very famous joke problem on the topic of permutations:
One day, 8 friends went into a restaurant in which there was a large round table, and argued for a long time among themselves about how best to sit around this table. They argued and argued until, finally, the owner of the restaurant offered them a deal: “Why are you arguing? None of you will remain hungry anyway :) First, sit down somehow! Remember today's seating arrangement well. Then come tomorrow and sit down differently. The next day come and sit down again in a new way! And so on... As soon as you go through all the possible seating options and it’s time to sit down again as you did today, then so be it, I promise to feed you in my restaurant for free!” Who will win – the owner or the visitors? :)
Well, let's count the number of everyone possible options seating arrangements. In our case, this is the number of permutations of 8 elements:
P 8 = 8! = 40320 ways.
Let us have 365 days in a year (we will not take into account leap days for simplicity). This means that even taking into account this assumption, the number of years it will take to try everything possible ways landings will be:
Over 110 years! That is, even if our heroes in wheelchairs are brought to the restaurant by their mothers straight from the maternity hospital, they will only be able to receive their free lunches at the age of very old centenarians. If, of course, all eight survive to that age.))
This is because factorial is a very fast increasing function! See for yourself:
By the way, what do the equalities and1! = 1 ? Here's how: from an empty set (0 elements) we can only create one permutation – empty set. :) Just like from a set consisting of just one element, we can also make only one permutation - this element itself.
Is everything clear with the rearrangements? Great, then let's do the tasks.)
Exercise 1
Calculate:
A)P 3 b)P5
IN)P 9:P 8 G)P2000:P1999
Task 2
Is it true that
Task 3
How many different four-digit numbers can be formed?
a) from the numbers 1, 2, 3, 4
b) from the numbers 0, 5, 6, 7?
Hint for point b): the number cannot begin with the number 0!
Task 4
Words and phrases with rearranged letters are called anagrams. How many anagrams can be made from the word "hypotenuse"?
Task 5
How many five-digit numbers divisible by 4 can be made by swapping the digits in the number 61135?
Hint: remember the test for divisibility by 4 (based on the last two digits)!
Answers in disarray: 2000; 3628800; 9; 24; 120; 18; 12; 6.
Well, everything worked out! Congratulations! Level 1 is completed, let's move on to the next one. Called " Placements without repetition."
FACTORIAL.
Factorial – this is the name of a function often encountered in practice, defined for non-negative integers. The name of the function comes from the English mathematical term factor- “multiplier”. It is designated n!. Factorial sign " ! "was introduced in 1808 in the French textbook Chr. Krump.
For every positive integer n function n! equal to the product of all integers from 1 before n.
For example:
4! = 1*2*3*4 = 24.
For convenience, we assume by definition 0! = 1 . The fact that the zero factorial must, by definition, be equal to one, was written in 1656 by J. Wallis in “The Arithmetic of the Infinite.”
Function n! grows with increasing n very fast. So,
(n+1)! = (n + 1) n! = (n + 1) n (n – 1)! (1)
English mathematician J. Stirling in 1970 offered a very convenient formula for approximate calculation of the function n!:
Where e = 2.7182... is the base of natural logarithms.
The relative error when using this formula is very small and falls quickly as the number n increases.
Let's look at ways to solve expressions containing factorial using examples.
Example 1. (n! + 1)! = (n! + 1) n! .
Example 2. Calculate 10! 8!
Solution. Let's use formula (1):
10! = 10*9*8! = 10*9=90 8! 8!
Example 3. Solve the equation (n + 3)! = 90 (n+1)!
Solution. According to formula (1) we have
= (n + 3)(n + 2) = 90.
(n + 3)! = (n + 3)(n + 2)(n+1)!(n+1)! (n+1)!
Opening the brackets in the product, we obtain a quadratic equation
n 2 + 5n - 84 = 0, whose roots are the numbers n = 7 and n = -12. However, the factorial is defined only for non-negative integers, that is, for all integers n ≥ 0. Therefore, the number n = -12 does not satisfy the conditions of the problem. So n = 7.
Example 4. Find at least one triple of natural numbers x, y and z, for which the equality x! = y! z!.
Solution. From the definition of the factorial of a natural number n it follows that
(n+1)! = (n + 1) n!
Let us put n + 1 = y in this equality! = x, Where at is an arbitrary natural number, we get
Now we see that the required triples of numbers can be specified in the form
(y!;y;y!-1) (2)
where y is a natural number greater than 1.
For example, the equalities are true
Example 5. Determine how many zeros end in the decimal notation of the number 32!.
Solution. If the decimal notation of a number R= 32! ends k zeros, then the number R can be represented in the form
P = q 10k,
where is the number q is not divisible by 10. This means that the decomposition of a number q prime factors does not contain both 2 and 5.
Therefore, to answer the question posed, let’s try to determine with what exponents the product 1 2 3 4 ... 30 31 32 includes the numbers 2 and 5. If the number k- the smallest of the found indicators, then the number P will end k zeros.
So, let’s determine how many numbers among the natural numbers from 1 to 32 are divisible by 2. Obviously, their number is 32/2 = 16. Then we’ll determine how many of the 16 numbers found are divisible by 4; then - how many of them are divisible by 8, etc. As a result, we get that among the first thirty-two natural numbers, 16 numbers are divisible by 2,
of which 32/4 = 8 numbers are divisible by 4, of which 32/8 = 4 numbers are divisible by 8, of which 32/16 = 2 numbers are divisible by 16, and finally, of these 32/32 = 1 are divisible by 32, those. one number. It is clear that the sum of the quantities received:
16 + 8 + 4 + 2 + 1 = 31
equal to the exponent with which the number 2 is included in 32!.
Similarly, let’s determine how many numbers among the natural numbers from 1 to 32 are divisible by 5, and from the found number by 10. Divide 32 by 5.
We get 32/5 = 6.4. Therefore, among the natural numbers from 1 to 32
there are 6 numbers that are divisible by 5. One of them is divisible by 25
number, since 32/25 = 1.28. As a result, the number 5 is included in the number 32! with an indicator equal to the sum 6+1 = 7.
From the results obtained it follows that 32!= 2 31 5 7 T, where is the number T is not divisible by either 2 or 5. Therefore, the number is 32! contains a multiplier
10 7 and, therefore, ends in 7 zeros.
So, in this abstract the concept of factorial is defined.
The formula of the English mathematician J. Stirling for the approximate calculation of the function n is given!
When transforming expressions containing factorial, it is useful to use the equality
(n+1)! = (n + 1) n! = (n + 1) n (n – 1)!
Methods for solving problems with factorial are discussed in detail using examples.
Factorial is used in various formulas in combinatorics, in the ranks, etc.
For example, the number of ways to build n schoolchildren in one line equals n!.
Number n! equals, for example, the number of ways in which n different books can be arranged on a bookshelf, or, for example, the number 5! equal to the number of ways in which five people can be seated on one bench. Or, for example, the number 27! equal to the number of ways our class of 27 students can line up in a physical education class.
Literature.
Ryazanovsky A.R., Zaitsev E.A.
Mathematics. 5-11 grades: Additional materials for a math lesson. –M.: Bustard, 2001.- (Teacher’s Library).
Encyclopedic dictionary of a young mathematician. / Comp. A.P.Savin.-M.: Pedagogy, 1985
Mathematics. School Student's Handbook. / Comp. G.M. Yakusheva.- M.: Philologist. society "Slovo", 1996.
What are factorials and how to solve them
The factorial of a number n, which in mathematics is denoted by the Latin letter n followed by an exclamation mark!. This expression is pronounced by voice as “n factorial”. A factorial is the result of sequential multiplication of a sequence of natural numbers from 1 to the desired number n. For example, 5! = 1 x 2 x 3 x 4 x 5 = 720 The factorial of the number n is denoted by the Latin letter n! and is pronounced en factorial. Represents the sequential multiplication (product) of all natural numbers starting from 1 to the number n. For example: 6! = 1 x 2 x 3 x 4 x 5=720
A factorial has a mathematical meaning only if the number is an integer and positive (natural). This meaning follows from the very definition of factorial, because All natural numbers are non-negative and integers. The values of factorials, namely the result of multiplying a sequence from one to the number n, can be viewed in the table of factorials. Such a table is possible because the factorial value of any integer is known in advance and is, so to speak, a table value.
By definition 0! = 1. That is, if there is a zero factorial, then we do not multiply anything and the result will be the first natural number that exists, that is, one.
The growth of the factorial function can be displayed on a graph. This will be an arc similar to the x-squared function, which will tend quickly upward.
Factorial is a fast growing function. It grows according to the graph faster than a polynomial function of any degree and even an exponential function. The factorial grows faster than a polynomial of any degree and an exponential function (but at the same time slower than a double exponential function). That is why it can be difficult to calculate the factorial manually, since the result can be very big number. To avoid calculating the factorial manually, you can use a factorial calculator, with which you can quickly get the answer. The factorial is used in functional analysis, number theory and combinatorics, in which it has a great mathematical meaning associated with the number of all possible unordered combinations of objects (numbers).
Free online factorial calculator
Our free solver allows you to calculate factorials online of any complexity in a matter of seconds. All you need to do is simply enter your data into the calculator. You can also find out how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group.
FACTORIAL.
Factorial – this is the name of a function often encountered in practice, defined for non-negative integers. The name of the function comes from the English mathematical term factor- “multiplier”. It is designated n!. Factorial sign " ! "was introduced in 1808 in the French textbook Chr. Krump.
For every positive integer n function n! equal to the product of all integers from 1 before n.
For example:
4! = 1*2*3*4 = 24.
For convenience, we assume by definition 0! = 1 . The fact that the zero factorial must, by definition, be equal to one, was written in 1656 by J. Wallis in “The Arithmetic of the Infinite.”
Function n! grows with increasing n very fast. So,
(n+1)! = (n + 1) n! = (n + 1) n (n – 1)! (1)
English mathematician J. Stirling in 1970 offered a very convenient formula for approximate calculation of the function n!:
Where e = 2.7182... is the base of natural logarithms.
The relative error when using this formula is very small and falls quickly as the number n increases.
Let's look at ways to solve expressions containing factorial using examples.
Example 1. (n! + 1)! = (n! + 1) n! .
Example 2. Calculate 10! 8!
Solution. Let's use formula (1):
10! = 10*9*8! = 10*9=90 8! 8!
Example 3. Solve the equation (n + 3)! = 90 (n+1)!
Solution. According to formula (1) we have
= (n + 3)(n + 2) = 90.
(n + 3)! = (n + 3)(n + 2)(n+1)!(n+1)! (n+1)!
Opening the brackets in the product, we obtain a quadratic equation
n 2 + 5n - 84 = 0, whose roots are the numbers n = 7 and n = -12. However, the factorial is defined only for non-negative integers, that is, for all integers n ≥ 0. Therefore, the number n = -12 does not satisfy the conditions of the problem. So n = 7.
Example 4. Find at least one triple of natural numbers x, y and z, for which the equality x! = y! z!.
Solution. From the definition of the factorial of a natural number n it follows that
(n+1)! = (n + 1) n!
Let us put n + 1 = y in this equality! = x, Where at is an arbitrary natural number, we get
Now we see that the required triples of numbers can be specified in the form
(y!;y;y!-1) (2)
where y is a natural number greater than 1.
For example, the equalities are true
Example 5. Determine how many zeros end in the decimal notation of the number 32!.
Solution. If the decimal notation of a number R= 32! ends k zeros, then the number R can be represented in the form
P = q 10k,
where is the number q is not divisible by 10. This means that the decomposition of a number q prime factors does not contain both 2 and 5.
Therefore, to answer the question posed, let’s try to determine with what exponents the product 1 2 3 4 ... 30 31 32 includes the numbers 2 and 5. If the number k- the smallest of the found indicators, then the number P will end k zeros.
So, let’s determine how many numbers among the natural numbers from 1 to 32 are divisible by 2. Obviously, their number is 32/2 = 16. Then we’ll determine how many of the 16 numbers found are divisible by 4; then - how many of them are divisible by 8, etc. As a result, we get that among the first thirty-two natural numbers, 16 numbers are divisible by 2,
of which 32/4 = 8 numbers are divisible by 4, of which 32/8 = 4 numbers are divisible by 8, of which 32/16 = 2 numbers are divisible by 16, and finally, of these 32/32 = 1 are divisible by 32, those. one number. It is clear that the sum of the quantities received:
16 + 8 + 4 + 2 + 1 = 31
equal to the exponent with which the number 2 is included in 32!.
Similarly, let’s determine how many numbers among the natural numbers from 1 to 32 are divisible by 5, and from the found number by 10. Divide 32 by 5.
We get 32/5 = 6.4. Therefore, among the natural numbers from 1 to 32
there are 6 numbers that are divisible by 5. One of them is divisible by 25
number, since 32/25 = 1.28. As a result, the number 5 is included in the number 32! with an indicator equal to the sum 6+1 = 7.
From the results obtained it follows that 32!= 2 31 5 7 T, where is the number T is not divisible by either 2 or 5. Therefore, the number is 32! contains a multiplier
10 7 and, therefore, ends in 7 zeros.
So, in this abstract the concept of factorial is defined.
The formula of the English mathematician J. Stirling for the approximate calculation of the function n is given!
When transforming expressions containing factorial, it is useful to use the equality
(n+1)! = (n + 1) n! = (n + 1) n (n – 1)!
Methods for solving problems with factorial are discussed in detail using examples.
Factorial is used in various formulas in combinatorics, in the ranks, etc.
For example, the number of ways to build n schoolchildren in one line equals n!.
Number n! equals, for example, the number of ways in which n different books can be arranged on a bookshelf, or, for example, the number 5! equal to the number of ways in which five people can be seated on one bench. Or, for example, the number 27! equal to the number of ways our class of 27 students can line up in a physical education class.
Literature.
Ryazanovsky A.R., Zaitsev E.A.
Mathematics. 5-11 grades: Additional materials for the mathematics lesson. –M.: Bustard, 2001.- (Teacher’s Library).
Encyclopedic dictionary of a young mathematician. / Comp. A.P.Savin.-M.: Pedagogy, 1985
Mathematics. School Student's Handbook. / Comp. G.M. Yakusheva.- M.: Philologist. society "Slovo", 1996.
The query reminds why a number raised to zero power is one, a query I resolved in an earlier article. Moreover, let me assure what I previously assured in explaining this obvious, shamelessly accepted, but inexplicable fact - the relationship is not arbitrary.
There are three ways to determine why the factor zero is equal to one.
Complete the template
1! = 1 * 1 = 1
2! = 1 * 2 = 2
3! = 1 * 2 * 3 = 6
4! = 1 * 2 * 3 * 4 = 24
If, (n-1)! = 1 * 2 * 3 * 4
,(P-3) * (n-2) * (N-1)
Then, logically, n! = 1 * 2 * 3 * 4
,(P-3) * (p-2) * (p-1) * p
Or, n! = n * (n-1)! - (i)
If you look closely at these trails, the picture reveals itself. Let's terminate it until it manages to produce legitimate results:
4! / 4 = 3!
3! / 3 = 2!
2! / 2 = 1!
1! / 1 = 0!
Or, 0! = 1
One can arrive at this result by simply plugging in the 1 for "n" in (i) to get:
1! = 1 * (1-1)!
1 = 1 * 0!
Or, 0! = 1
However, this explanation says nothing about why factorials of negative numbers cannot exist. Let's look at our pattern again to find out why.
2! / 2 = 1!
1! / 1 = 0!
0! / 0 =
,I would agree that these methods are a bit suspect; they seem to be crafty, implicit ways of defining the factorial of zero. It's like arguing for straw. However, one can find an explanation in a field whose entire existence depends on the calculation of factorials - combinatorics.
Agreements
Consider 4 chairs that must be occupied by 4 people. The first chair could be occupied by any of these four people, so the resulting number of choices would be 4. Now that one chair is occupied, we have 3 options that could potentially be occupied for the next chair. Likewise, the next chair represents two options, and the last chair represents one choice; he is occupied by the last person. Thus, the total number of selections we have is 4x3x2x1 or 4!. Or you could say there are 4! ways to organize 4 different chairs.
So when the value of "n" is zero, the question turns to what are the different ways to organize zero number of objects? One, of course! There is only one permutation or one way to arrange nothing, because there is nothing to arrange. WHAT? To be fair, it belongs to a branch of philosophy, albeit one of the nasty or false ideas that freshmen trust after reading Nietzsche quotes on Pinterest.
Let's look at an example that involves physical objects, as this may improve understanding. Factorials are also central to computer combinations, a process that also determines mechanisms, but unlike permutation, the order of things does not matter. The difference between permutation and combination is the difference between a combination lock and a bowl of fruit cubes. Combination locks are often mistakenly called "combination locks" when they are actually called permutations, since 123 and 321 cannot unlock them.
The general formula for determining the number of paths of "k" objects can be arranged among "n" places:
Whereas, to determine the number of ways to select or combine "k" objects from "n" objects:
This allows us to, say, determine the number of ways in which two balls can be selected from a bag that contains five balls of different colors. Since the order of the selected balls is not important, we refer to the second formula to calculate the attracting combinations.
So what if the values of "n" and "k" are exactly the same? Let's replace these values and find out. Note that the factorial of zero is obtained in the denominator.
But how do we understand this mathematical calculation visually, from the point of view of our example? The calculation is essentially a solution to a question that asks: What are the different number of ways in which we can select three balls from a bag containing only three balls? Well, of course! Selecting them in any order will have no effect! The calculation equation with one and factorial zero turns out to be *drum roll*
..
