Combinatorics is, as the name itself hints, 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, it is difficult to understand. Why? Because it often contains terms and symbols 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!,P9 . . problems begin. In addition, in most textbooks, this topic is presented rather dryly and difficult to understand. I hope this material will help at least a little to solve these problems 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 which combination of products is chosen - tasty or tasteless. True, it is no longer mathematics that deals with taste issues, but cooking, but nevertheless.) When we play “words”, composing small words from one long one, we combine letters. When we open the combination lock or dial a phone number, we combine the numbers.) The head teacher of the school draws up lesson schedules by combining subjects. Football teams at the World or European Championships are divided into groups, forming combinations. And so on.)
Combinatorial problems were solved by people in ancient times ( magic squares, chess), and the real flowering of combinatorics came in the 6th-7th centuries, during the wide spread of gambling (cards, dice), when players had to think through various moves and thereby actually also solve combinatorial problems.) Together with combinatorics, this time, another branch of mathematics was born - probability theory . These two sections are very close relatives and go hand in hand.) And when studying probability theory, we will often encounter problems of combinatorics.
And we will begin the study of combinatorics with such a cornerstone concept as factorial .
What is factorial?
The beautiful word "factorial", but frightens and confuses many. But in vain. In this lesson, we will understand and work hard 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 a factorial.
Let's take some natural number n . Completely arbitrary: we want 2, we want 10, - whatever, as long as it is natural.) So, factorial of a natural number n is the work of all natural numbers from 1 to n inclusive. It is marked like this: n! That is,
In order not to paint it every time long work, just came up with a short notation. :) It is read a little unusually: “en factorial” (and not vice versa “factorial en”, as it may seem).
And that's it! For example,
Do you get the idea?)) Great! Then consider examples:
Answers (in disarray): 30; 0.1; 144; 6; 720; 2; 5040.
Everything worked out? Wonderful! We already know how to count factorials and solve simple examples with them. Go ahead. :)
Factorial Properties
Consider the expression 0, which is not very clear from the point of view of the definition of the factorial! So it is agreed in mathematics that
Yes Yes! This is an interesting equality. What is from one, what is 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 properties are very similar:
They are proved elementarily. Just like the factorial.)
These two formulas allow, firstly, to easily calculate the factorial of the current natural number through the factorial previous numbers. Or next 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. Kind of like that.
Calculate:
How will we act? Successively multiply all natural numbers from 1 to 1999 and from 1 to 2000? It's crazy! But according to the properties, the example is 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 - a philosophical question. In mathematics, just like that, purely for beauty, nothing happens.)) In fact, the 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 such an interesting infinite sum:
The more askedn , topics more terms in the sum and the closer this sum will be to the numbere . And in limit when it becomes exactly equal to the numbere . :) But we will talk about this amazing number in the corresponding topic. And here we have factorials and combinatorics.)
Where did they come from? They came just from combinatorics, from the study of sets of elements.) The simplest such set is permutation without repetition. Let's start with her. :)
Permutation without repetition
Suppose we 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 be done with them? If these are sweets, then, of course, you can eat them.)) For now, we will tolerate them and we will arrange in a different order.
Each such arrangement is called permutation without repetition. Why "no repetition"? Because all the elements involved in the permutation different. This is what we have decided so far for simplicity. Is there some more permutation with repetition, where some elements may be the same. But such permutations are a bit 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 set.C . In this case, there will be six permutations:
ABC , ACB , BAC , BCA , CAB , CBA .
Permutations of four elements will be constructed as follows. First, put the element in the first placeA . At the same time, the remaining three element can be rearranged, as we already know, six ways:
So the number of permutations with the first elementA equals 6.
But the same story will turn out if we put in the first place any of these four elements. They are equal and everyone deserves to be in first place.) So, the total number of permutations of the four elements will be equal to . Here they are:
So let's recap: permutation from n elements is called any ordered a set of these nelements .
The word "ordered" is key here: each permutation differs only element order, while the elements 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, so every time we write out All different options and count them. :) For 4 elements, we got 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, we formulate one auxiliary rule, which is very important in all combinatorics, called product rule .
Product rule: if the set contains n various options selection of the first element and for each of them there is m different options for choosing the second element, then the total can be composed n m different pairs of these elements.
Now, let's now have a set ofn various elements
,
where, of course, . We need to count the number of all possible permutations from the elements of this set. We argue in exactly the same way.)) In the first place, you can put any of thesen 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 (because 2 items 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 have already been selected in one way or another. :)
Well, now we construct the formula.
So, the number of ways to select the first element from the set isn . On every of thesen ways ton-1 way to choose the second. This means that the total number of ways to select the 1st and 2nd elements, according to product rule, will be equal ton(n-1) . Further, each of them, in turn, accounts forn-2 way to select the third element. Means, three element can 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 chosen (or in our case arranged) ways.
The number of such ways is indicated as follows:P n . It reads: “pe from en”. From French " P ermutation - permutation". Translated into Russian means: "permutation from n elements".
Means,
Now let's look at the expressionon the right side of the formula. Doesn't it remind you of anything? And if you rewrite from right to left, like this?
Well, of course! Factorial, in person. :) Now we can briefly write:
Means, number all possible permutations from n different elements n! .
This is the main practical meaning of the 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) of 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. :)
It's great, right?
On the topic of permutations, there is one very well-known joke problem:
Once 8 friends went into a restaurant where there was a large round table, and argued among themselves for a long time about how best they should sit around this table. They argued and argued until, finally, the owner of the restaurant offered them a deal: “What are you arguing about? None of you will remain hungry anyway :) Sit down for a start somehow! Remember today's seating well. Then come back tomorrow and sit down differently. The next day, come and sit 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 the way you did today, then so be it, I promise you to feed in my restaurant for free! Who will win - the owner or the visitors? :)
Well, let's count the number of all options seating arrangements. In our case, this is the number of permutations of 8 elements:
P8 = 8! = 40320 ways.
Let us have 365 days in a year (we will not take into account leap years for simplicity). So, even with this assumption, the number of years it will take to try all possible landing methods will be:
Over 110 years! That is, even if our heroes in wheelchairs are brought to the restaurant by their mothers directly from the hospital, they will be able to receive their free meals only at the age of very elderly centenarians. Unless, of course, all eight live to that age.))
This is because the factorial is a very fast growing function! See for yourself:
Incidentally, what do the equalities and1! = 1 ? And here's how: from an empty set (0 elements), we can only compose one permutation is an empty set. :) Just like from a set consisting of only one element, we can also compose only one permutation - this element itself.
Is everything clear with the permutations? Great, then let's do the tasks.)
Exercise 1
Calculate:
A)P3 b)P5
IN)P9:P8 G)P 2000:P 1999
Task 2
Is it true that
Task 3
How many different 4-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 start with the number 0!
Task 4
Words and phrases with rearranged letters are called anagrams. How many anagrams can you make with hypotenuse?
Task 5
How many five-digit numbers divisible by 4 can be formed by swapping the digits in the number 61135?
Hint: remember the sign of divisibility by 4 (by the last two digits)!
Responses in disarray: 2000; 3628800; 9; 24; 120; 18; 12; 6.
Well, everything worked out! Congratulations! Level 1 completed, move on to the next. called " Placements without repetition."
FACTORIAL.
Factorial - this is the name of a function often encountered in practice, defined for integers not negative numbers. 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! is 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 zero - factorial should be, by definition, equal to one, was written in 1656 by J. Vallis in Arithmetic of the Infinite.
Function n! grows with the increase n very fast. So,
(n + 1)! = (n + 1) n! = (n + 1) n (n - 1)! (1)
English mathematician J. Stirling in 1970 offered a very comfortable formula for an 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 off rapidly as the number n increases.
We will consider the ways of solving expressions containing factorial using examples.
Example 1. (n! + 1)! = (n! + 1) n! .
Example 2. Calculate 10! 8!
Solution. We 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)!
Expanding 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, i.e., for all integers n ≥ 0. Therefore, the number n = -12 does not satisfy the condition 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. It follows from the definition of the factorial of a natural number n that
(n+1)! = (n + 1) n!
We put in this equality n + 1 = y! = x, Where at is an arbitrary natural number, we obtain
Now we see that the desired 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
Example 5 Determine how many zeros the decimal representation of the number 32! ends with.
Solution. If the decimal notation R= 32! ends k zeros, then the number R can be represented as
P = q 10 k ,
where number q is not divisible by 10. This means that the expansion of the number q into prime factors does not contain both 2 and 5.
Therefore, in order to answer the question posed, let us try to determine with what indicators 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 natural numbers from 1 to 32 are divisible by 2. Obviously, their number is 32/2 = 16. Then we 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 thirty-two first 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 which 32/32=1 are divisible by 32, those. one number. It is clear that the sum of the received quantities:
16 + 8 + 4 + 2 + 1 = 31
is equal to the exponent with which the number 2 is included in 32!.
Similarly, we determine how many numbers among natural numbers from 1 to 32 are divisible by 5, and from the number found by 10. Divide 32 by 5.
We get 32/5 = 6.4. Therefore, among natural numbers from 1 to 32
there are 6 numbers that are divisible by 5. Of these, one 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.
It follows from the obtained results that 32!= 2 31 5 7 T, where number T is not divisible by 2 or 5. Therefore, the number 32! contains a multiplier
10 7 and, therefore, ends with 7 zeros.
So, in this essay the concept of factorial is defined.
The formula of the English mathematician J. Stirling for the approximate calculation of the function n!
When converting expressions containing a factorial, it is useful to use the equality
(n + 1)! = (n + 1) n! = (n + 1) n (n - 1)!
The examples show in detail how to solve problems with factorial.
The factorial is used in various formulas in combinatorics, in the ranks, etc.
For example, the number of ways to line up 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! is equal to the number of ways in which five people can be seated on one bench. Or, for example, the number 27! equals the number of ways our class of 27 students can line up in PE.
Literature.
Ryazanovsky A.R., Zaitsev E.A.
Mathematics. 5-11 cells: Additional materials to the math lesson. -M.: Bustard, 2001.- (Teacher's Library).
Encyclopedic Dictionary of a Young Mathematician. /Comp. A.P. Savin.-M.: Pedagogy, 1985
Mathematics. Student's handbook. /Comp. G.M. Yakusheva.- M.: Philologist. Society "Slovo", 1996.
What are factorials and how to solve them
The factorial of the 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! \u003d 1 x 2 x 3 x 4 x 5 \u003d 720 The factorial of the number n is denoted by the Latin letter n! and is pronounced en factorial. It is a sequential multiplication (product) of all natural numbers starting from 1 to the number n. For example: 6! \u003d 1 x 2 x 3 x 4 x 5 \u003d 720
The factorial has mathematical meaning only when if this number is integer and positive (natural). This meaning follows from the very definition of the factorial, since All natural numbers are non-negative and integer. The values of factorials, namely the result of multiplying a sequence from one to the number n, can be viewed in the factorial table. Such a table is possible, due to the fact that the value of the factorial of any integer is known in advance and is, so to speak, a tabular 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 existing number, that is one.
The growth of the factorial function can be plotted. It will be an arc, similar to the x-squared function, which will quickly go up.
Factorial is a fast growing function. It grows 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 slower than a double exponential function). That is why it can be difficult to calculate the factorial by hand, as the result can be a very large number. In order not to calculate the factorial manually, you can use the 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 seconds. All you need to do is just enter your data into the calculator. You can also learn how to solve the equation on our website. And if you have any 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! is 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 zero - factorial should be, by definition, equal to one, was written in 1656 by J. Vallis in Arithmetic of the Infinite.
Function n! grows with the increase n very fast. So,
(n + 1)! = (n + 1) n! = (n + 1) n (n - 1)! (1)
English mathematician J. Stirling in 1970 offered a very comfortable formula for an 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 off rapidly as the number n increases.
We will consider the ways of solving expressions containing factorial using examples.
Example 1. (n! + 1)! = (n! + 1) n! .
Example 2. Calculate 10! 8!
Solution. We 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)!
Expanding 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, i.e., for all integers n ≥ 0. Therefore, the number n = -12 does not satisfy the condition 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. It follows from the definition of the factorial of a natural number n that
(n+1)! = (n + 1) n!
We put in this equality n + 1 = y! = x, Where at is an arbitrary natural number, we obtain
Now we see that the desired 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
Example 5 Determine how many zeros the decimal representation of the number 32! ends with.
Solution. If the decimal notation R= 32! ends k zeros, then the number R can be represented as
P = q 10 k ,
where number q is not divisible by 10. This means that the expansion of the number q into prime factors does not contain both 2 and 5.
Therefore, in order to answer the question posed, let us try to determine with what indicators 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 natural numbers from 1 to 32 are divisible by 2. Obviously, their number is 32/2 = 16. Then we 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 thirty-two first 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 which 32/32=1 are divisible by 32, those. one number. It is clear that the sum of the received quantities:
16 + 8 + 4 + 2 + 1 = 31
is equal to the exponent with which the number 2 is included in 32!.
Similarly, we determine how many numbers among natural numbers from 1 to 32 are divisible by 5, and from the number found by 10. Divide 32 by 5.
We get 32/5 = 6.4. Therefore, among natural numbers from 1 to 32
there are 6 numbers that are divisible by 5. Of these, one 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.
It follows from the obtained results that 32!= 2 31 5 7 T, where number T is not divisible by 2 or 5. Therefore, the number 32! contains a multiplier
10 7 and, therefore, ends with 7 zeros.
So, in this essay the concept of factorial is defined.
The formula of the English mathematician J. Stirling for the approximate calculation of the function n!
When converting expressions containing a factorial, it is useful to use the equality
(n + 1)! = (n + 1) n! = (n + 1) n (n - 1)!
The examples show in detail how to solve problems with factorial.
The factorial is used in various formulas in combinatorics, in the ranks, etc.
For example, the number of ways to line up 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! is equal to the number of ways in which five people can be seated on one bench. Or, for example, the number 27! equals the number of ways our class of 27 students can line up in PE.
Literature.
Ryazanovsky A.R., Zaitsev E.A.
Mathematics. Grades 5-11: Additional materials for the lesson of mathematics. -M.: Bustard, 2001.- (Teacher's Library).
Encyclopedic Dictionary of a Young Mathematician. /Comp. A.P. Savin.-M.: Pedagogy, 1985
Mathematics. Student's handbook. /Comp. G.M. Yakusheva.- M.: Philologist. Society "Slovo", 1996.
The query is reminiscent of why a number raised to zero power is one, a query I resolved in an earlier article. Also, let me reassure what I have previously reassured when explaining this obvious, shamelessly accepted but inexplicable fact - the attitude is not arbitrary.
There are three ways to determine why a factor zero is equal to one.
Complete 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) * (P-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 paths, the picture will show itself. Let's complete it until it manages to get 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 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 see why.
2! / 2 = 1!
1! / 1 = 0!
0! / 0 =
,I would agree that these methods are a little suspicious; 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 the field, its entire existence depends on the calculation of factorials - combinatorics.
Arrangements
Consider 4 chairs to be occupied by 4 people. The first chair can be occupied by any of these four people, so the resulting number of choices is 4. Now that one chair is occupied, we have 3 options that could potentially be occupied for the next chair. Similarly, the next chair represents two choices, and the last chair represents one choice; He is busy the last man. So the total number of choices 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 translates to what are the different ways to organize the 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, this belongs to a branch of philosophy, albeit one of the nasty or false notions of what freshmen trust after reading Nietzsche's Pinterest quotes.
Let's look at an example that includes 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 doesn't matter. The difference between permutation and combination lies in the difference between a combination lock and a bowl of fruit cube melange. Combination locks are often erroneously referred to as "combination locks" when they are actually called permutations, as 123 and 321 cannot unlock them.
The general formula for determining the number of paths of "k" objects can be organized 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 marbles can be selected from a bag that contains five balls of different colors. Since the order of the chosen balls is not important, we refer to the second formula for calculating the drawing 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, in terms of our example? The calculation is essentially a solution to a question that asks: what are the different number of ways we can choose three marbles from a bag containing only three marbles? Well, of course! Selecting them in any order will not affect! Calculation equation with one and factorial of zero turns out to be *drum roll*
..
