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– How to Master Combinatorics: A PDF Guide on Permutations, Variations and Combinations
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# How to Master Combinatorics: A PDF Guide on Permutations, Variations and Combinations

Combinatorics is a branch of mathematics that deals with the study of finite or countable discrete structures. It involves counting, arranging, selecting and grouping objects according to certain rules or criteria. Combinatorics has many applications in fields such as cryptography, coding theory, graph theory, probability, statistics and computer science.

One of the most important concepts in combinatorics is the notion of permutation, variation and combination. These are different ways of forming subsets or sequences from a given set of elements. In this PDF guide, we will explain the definitions, formulas and examples of these concepts, as well as some useful properties and techniques for solving combinatorial problems.

We will assume that you have some basic knowledge of algebra and arithmetic, as well as the factorial notation (n! = n · (n-1) · … · 2 · 1). We will also use the binomial coefficient notation (n choose k), which represents the number of ways to choose k elements from a set of n elements. The binomial coefficient can be calculated by the formula:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

We will use the following symbols and conventions throughout this guide:

– n: the number of elements in a given set
– k: the number of elements in a subset or a sequence
– S: a set of n elements
– P: a permutation of k elements from S
– V: a variation of k elements from S
– C: a combination of k elements from S
– P(n): the number of permutations of n elements
– V(n,k): the number of variations of k elements from n elements
– C(n,k): the number of combinations of k elements from n elements

We will also distinguish between two types of subsets or sequences: those with repetition and those without repetition. Repetition means that an element can appear more than once in a subset or a sequence. For example, (a,a,b) is a sequence with repetition, while (a,b,c) is a sequence without repetition.

In the next sections, we will define and illustrate each concept in detail.

## Combination

A combination of k elements from a set S of n elements is an unordered selection of k distinct elements from S. For example, if S = {a, b, c, d}, then {a, b, c} and {c, d, a} are combinations of 3 elements from S.

The number of combinations of k elements from n elements is denoted by C(n,k) and can be calculated by the formula:

$$C(n,k) = \frac{n!}{k!(n-k)!} = \binom{n}{k}$$

The formula can be understood as follows: to form a combination of k elements from n elements, we can first choose a permutation of k elements from n elements (which has P(n,k) ways), and then divide by the number of permutations of k elements (which has k! ways), since the order does not matter.

For example, if we want to find the number of combinations of 3 letters from the alphabet (n = 26), we can use the formula:

$$C(26,3) = \frac{26!}{3!(26-3)!} = \binom{26}{3} = 2600$$

This means that there are 2600 different ways to choose 3 letters from the alphabet without order.

A special case of combination is when k = n, that is, when we want to choose all the elements of a set without order. This is called a combination of n elements and is denoted by C(n). The formula for C(n) is simply:

$$C(n) = 1$$

This means that there is only one way to choose n elements from n elements without order.

For example, if we want to find the number of combinations of 4 digits without order (n = 10), we can use the formula:

$$C(10) = 1$$

This means that there is only one way to choose 10 digits without order.

## Binomial Theorem

The binomial theorem is a useful formula that allows us to expand a binomial expression raised to a positive integer power. A binomial expression is an algebraic expression that consists of two terms, such as (x+y) or (a-b). The binomial theorem states that:

$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$

The formula can be understood as follows: to expand (x+y)^n, we can consider all the possible combinations of x and y terms that have a total degree of n. For each combination, we need to multiply the coefficients, which are given by the binomial coefficients C(n,k), and the powers of x and y, which are given by x^(n-k) and y^k.

For example, if we want to expand (x+y)^3, we can use the binomial theorem:

$$(x+y)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3-k} y^k$$
$$= \binom{3}{0} x^3 y^0 + \binom{3}{1} x^2 y^1 + \binom{3}{2} x^1 y^2 + \binom{3}{3} x^0 y^3$$
$$= x^3 + 3x^2y + 3xy^2 + y^3$$

The binomial theorem can be used to solve many combinatorial problems involving binomial expressions. For example, if we want to find the coefficient of x^5 in the expansion of (x+2)^7, we can use the binomial theorem:

$$(x+2)^7 = \sum_{k=0}^7 \binom{7}{k} x^{7-k} (2)^k$$

The coefficient of x^5 is given by the term where k = 2:

$$\binom{7}{2} x^{7-2} (2)^2 = 21 x^5 (4) = 84 x^5$$

Therefore, the coefficient of x^5 is 84.

## Exercises

Here are some exercises to practice your skills on permutation, variation and combination. Try to solve them by yourself before checking the answers.

1. How many different words can be formed by rearranging the letters of the word MATHEMATICS?
2. How many different four-digit numbers can be formed by using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 if repetition is not allowed?
3. How many different ways can a committee of 5 people be chosen from a group of 10 people?
4. How many different ways can a hand of 5 cards be dealt from a standard deck of 52 cards?
5. What is the coefficient of x^7 in the expansion of (x+3)^10?

## Answers

1. The word MATHEMATICS has 11 letters, but some of them are repeated. There are 2 M’s, 2 A’s, 2 T’s and 2 I’s. The number of different words that can be formed by rearranging the letters is given by the formula:

$$P_{11}(2,2,2,2) = \frac{11!}{2!2!2!2!} = 4989600$$

2. To form a four-digit number, we need to choose one digit for each position from left to right. Since repetition is not allowed, we have 9 choices for the first digit (we cannot use 0), 8 choices for the second digit, 7 choices for the third digit and 6 choices for the fourth digit. The number of different four-digit numbers that can be formed is given by the formula:

$$V_4(9) = \frac{9!}{(9-4)!} = 9 \cdot 8 \cdot 7 \cdot 6 = 3024$$

3. To choose a committee of 5 people from a group of 10 people, we need to select 5 distinct elements from a set of 10 elements. The order of selection does not matter, so we use combination. The number of different ways to choose a committee is given by the formula:

$$C(10,5) = \frac{10!}{5!(10-5)!} = \binom{10}{5} = 252$$

4. To deal a hand of 5 cards from a standard deck of 52 cards, we need to select 5 distinct elements from a set of 52 elements. The order of selection does not matter, so we use combination. The number of different ways to deal a hand is given by the formula:

$$C(52,5) = \frac{52!}{5!(52-5)!} = \binom{52}{5} = 2598960$$

5. To find the coefficient of x^7 in the expansion of (x+3)^10, we need to use the binomial theorem:

$$(x+3)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^{10-k} (3)^k$$

The coefficient of x^7 is given by the term where k = 3:

$$\binom{10}{3} x^{10-3} (3)^3 = 120 x^7 (27) = 3240 x^7$$

Therefore, the coefficient of x^7 is 3240.

## Applications

Permutation, variation and combination are useful concepts in many fields of mathematics and science. Here are some examples of how they can be applied to solve real-world problems.

– Permutation can be used to count the number of possible arrangements of objects or symbols. For example, if we want to find the number of possible passwords that can be formed by using 8 alphanumeric characters (26 letters and 10 digits), we can use permutation:

$$P_{36}(8) = \frac{36!}{(36-8)!} = 36 \cdot 35 \cdot 34 \cdot 33 \cdot 32 \cdot 31 \cdot 30 \cdot 29 = 2176782336$$

This means that there are over 2 billion possible passwords that can be formed by using 8 alphanumeric characters.

– Variation can be used to count the number of possible selections of objects or symbols with a given order. For example, if we want to find the number of possible license plates that can be formed by using 3 letters followed by 4 digits, we can use variation:

$$V_3(26) \cdot V_4(10) = \frac{26!}{(26-3)!} \cdot \frac{10!}{(10-4)!} = 26 \cdot 25 \cdot 24 \cdot 10 \cdot 9 \cdot 8 \cdot 7 = 35880000$$

This means that there are over 35 million possible license plates that can be formed by using this format.

– Combination can be used to count the number of possible selections of objects or symbols without a given order. For example, if we want to find the number of possible poker hands (5 cards) that can be formed from a standard deck of 52 cards, we can use combination:

$$C(52,5) = \frac{52!}{5!(52-5)!} = \binom{52}{5} = 2598960$$

This means that there are over 2.5 million possible poker hands that can be formed from a standard deck of cards.

## Solutions

Here are some solutions to the exercises given in the previous section. You can check your answers and see the steps involved in solving them.

1. To form a word by rearranging the letters of MATHEMATICS, we need to choose one letter for each position from left to right. Since some letters are repeated, we have to divide by the number of ways to arrange those letters among themselves. The number of different words that can be formed is given by the formula:

$$P_{11}(2,2,2,2) = \frac{11!}{2!2!2!2!} = 4989600$$

$$V_4(9) = \frac{9!}{(9-4)!} = 9 \cdot 8 \cdot 7 \cdot 6 = 3024$$

$$C(10,5) = \frac{10!}{5!(10-5)!} = \binom{10}{5} = 252$$

$$C(52,5) = \frac{52!}{5!(52-5)!} = \binom{52}{5} = 2598960$$

5. To find the coefficient of x^7 in the expansion of (x+3)^10, we need to use the binomial theorem:

$$(x+3)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^{10-k} (3)^k$$

The coefficient of x^7 is given by the term where k = 3:

$$\binom{10}{3} x^{10-3} (3)^3 = 120 x^7 (27) = 3240 x^7$$

Therefore, the coefficient of x^7 is 3240.

## Conclusion

In this article, we have learned about the concepts of permutation, variation and combination, and how to use them to count the number of possible arrangements, selections and distributions of objects or symbols. We have also seen some examples of how these concepts can be applied to solve real-world problems involving passwords, license plates, poker hands and more. We have also practiced our skills on some exercises and solutions. We hope that this article has helped you to understand and appreciate the beauty and usefulness of combinatorics. Thank you for reading!

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