# Learn Permutations and Combinations with this Free PDF Tutorial - 14 Express Principales Explained

# Permutations and Combinations Tutorial PDF Free 14 Express Principales ## Introduction - Explain what permutations and combinations are and why they are important - Give some examples of problems that involve permutations and combinations - State the main objectives of the article ## Permutations - Define permutations and give the formula for calculating them - Explain the difference between permutations with and without repetition - Give some examples of how to use permutations to solve problems ## Combinations - Define combinations and give the formula for calculating them - Explain the difference between combinations with and without repetition - Give some examples of how to use combinations to solve problems ## Factorials - Define factorials and give the notation for them - Explain how factorials are related to permutations and combinations - Give some properties and rules of factorials ## Binomial Coefficients - Define binomial coefficients and give the notation for them - Explain how binomial coefficients are related to combinations - Give some properties and rules of binomial coefficients ## Pascal's Triangle - Define Pascal's triangle and explain how it is constructed - Explain how Pascal's triangle is related to binomial coefficients and combinations - Give some patterns and applications of Pascal's triangle ## The Binomial Theorem - State the binomial theorem and explain what it means - Explain how the binomial theorem is related to binomial coefficients and Pascal's triangle - Give some examples of how to use the binomial theorem to expand binomials ## The Multinomial Theorem - State the multinomial theorem and explain what it means - Explain how the multinomial theorem is a generalization of the binomial theorem - Give some examples of how to use the multinomial theorem to expand multinomials ## Permutations with Indistinguishable Objects - Explain what indistinguishable objects are and how they affect permutations - Give the formula for calculating permutations with indistinguishable objects - Give some examples of how to use permutations with indistinguishable objects to solve problems ## Combinations with Repetition - Explain what repetition is and how it affects combinations - Give the formula for calculating combinations with repetition - Give some examples of how to use combinations with repetition to solve problems ## The Inclusion-Exclusion Principle - State the inclusion-exclusion principle and explain what it means - Explain how the inclusion-exclusion principle is related to permutations and combinations - Give some examples of how to use the inclusion-exclusion principle to solve problems ## The Pigeonhole Principle - State the pigeonhole principle and explain what it means - Explain how the pigeonhole principle is related to permutations and combinations - Give some examples of how to use the pigeonhole principle to solve problems ## Permutations and Combinations in Probability - Explain what probability is and how it is calculated - Explain how permutations and combinations are used to calculate probabilities of events - Give some examples of how to use permutations and combinations in probability problems ## Permutations and Combinations in Statistics - Explain what statistics is and how it is used to analyze data - Explain how permutations and combinations are used to calculate statistical measures such as mean, variance, standard deviation, etc. - Give some examples of how to use permutations and combinations in statistics problems ## Conclusion - Summarize the main points of the article - Emphasize the importance and usefulness of permutations and combinations in mathematics and other fields - Provide some resources for further learning Now that I have created an outline, I will start writing the article based on it. Here is the article I have written: # Permutations And Combinations Tutorial PDF Free 14 Express Principales Permutations and combinations are two fundamental concepts in mathematics that deal with arranging or selecting objects from a set. They have many applications in various fields such as computer science, cryptography, combinatorics, probability, statistics, etc. In this article, we will learn what permutations and combinations are, how to calculate them, and how to use them to solve various problems. We will also explore some related topics such as factorials, binomial coefficients, Pascal's triangle, binomial theorem, multinomial theorem, inclusion-exclusion principle, pigeonhole principle, etc. By the end of this article, you should be able to: - Understand the difference between permutations and combinations and how to calculate them - Apply permutations and combinations to solve various problems involving arrangements and selections - Use factorials, binomial coefficients, Pascal's triangle, binomial theorem, multinomial theorem, inclusion-exclusion principle, pigeonhole principle, etc. to simplify calculations and find patterns - Use permutations and combinations in probability and statistics to calculate probabilities and statistical measures ## Permutations A permutation is an arrangement of objects in a specific order. For example, the letters A, B, C can be arranged in six different ways: ABC, ACB, BAC, BCA, CAB, CBA. Each of these arrangements is a permutation of the letters A, B, C. To calculate the number of permutations of n objects taken r at a time, we use the following formula: $$P(n,r) = \fracn!(n-r)!$$ where n! (read as n factorial) is the product of all positive integers from 1 to n. For example, $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$ The formula for permutations can be understood as follows: To arrange r objects from a set of n objects, we have n choices for the first object, n-1 choices for the second object, n-2 choices for the third object, and so on, until we have n-r+1 choices for the r-th object. Multiplying these choices together gives us the number of permutations. For example, to arrange 3 letters from a set of 5 letters (A, B, C, D, E), we have: $$P(5,3) = \frac5!(5-3)! = \frac5!2! = \frac5 \times 4 \times 3 \times 2 \times 12 \times 1 = 60$$ The formula for permutations assumes that the objects are distinct and that the order matters. If the objects are not distinct or if the order does not matter, then we need to modify the formula accordingly. For example, if we have a set of 4 letters (A, A, B, C) and we want to arrange them in different ways, we cannot use the formula P(4,4) = 24 because some of the arrangements are identical (e.g., AABB and AABB). To avoid counting duplicates, we need to divide by the number of ways to arrange the indistinguishable objects. In this case, we have two A's that can be arranged in 2! ways. Therefore, the number of permutations of 4 letters with two A's is: $$P(4;2) = \frac4!2! = \frac4 \times 3 \times 2 \times 12 \times 1 = 12$$ The notation P(4;2) means that we have 4 objects with two indistinguishable objects. Similarly, if we have a set of 5 letters (A, B, C, D, E) and we want to select and arrange 3 of them without regard to order (e.g., ABC and BAC are considered the same), then we cannot use the formula P(5,3) = 60 because some of the arrangements are equivalent. To avoid counting duplicates, we need to divide by the number of ways to arrange the selected objects. In this case, we have 3 objects that can be arranged in 3! ways. Therefore, the number of permutations of 5 letters taken 3 at a time without regard to order is: $$P(5;3) = \fracP(5;3)3! = \frac606 = 10$$ The notation P(5;3) means that we have 5 objects taken 3 at a time without regard to order. Here are some examples of how to use permutations to solve problems: - How many different ways can you arrange the letters in the word PERMUTATION? Solution: We have 11 letters with two A's and two T's. Therefore, $$P(11;2;2) = \frac11!2!2! = \frac11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 12 \times 1 \times 2 \times 1 = 19958400$$ - How many different four-digit numbers can be formed using the digits from 0;1;2;3;4;5;6;7;8;9? - Numbers that start with zero: We have 9 choices for the second digit (any digit except zero), 9 choices for the third digit (any digit), and 9 choices for the fourth digit (any digit). Therefore, the number of four-digit numbers that start with zero is: $$P(9;3) = 9 \times 9 \times 9 = 729$$ - Numbers that do not start with zero: We have 9 choices for the first digit (any digit except zero), 9 choices for the second digit (any digit), 8 choices for the third digit (any digit except the first one), and 7 choices for the fourth digit (any digit except the first two). Therefore, the number of four-digit numbers that do not start with zero is: $$P(9;1) \times P(9;1) \times P(8;1) \times P(7;1) = 9 \times 9 \times 8 \times 7 = 4536$$ - Adding these two cases together, we get the total number of four-digit numbers that can be formed using the digits from 0;1;2;3;4;5;6;7;8;9: $$729 + 4536 = 5265$$ - How many ways can a committee of 3 people be chosen from a group of 10 people and arranged in a row? Solution: We have two steps: choosing and arranging. To choose 3 people from a group of 10 people, we have: $$P(10;3) = \frac10!(10-3)! = \frac10!7! = 10 \times 9 \times 8 = 720$$ To arrange the chosen people in a row, we have: $$P(3;3) = \frac3!(3-3)! = \frac3!0! = 3! = 6$$ Multiplying these two steps together, we get the number of ways to choose and arrange a committee of 3 people from a group of 10 people: $$720 \times 6 = 4320$$ ## Combinations A combination is a selection of objects from a set without regard to order. For example, the letters A, B, C can be selected in four different ways: A, B, C, AB, AC, BC, ABC. Each of these selections is a combination of the letters A, B, C. To calculate the number of combinations of n objects taken r at a time, we use the following formula: $$C(n,r) = \fracn!r!(n-r)!$$ where n! and r! are factorials as defined before. The formula for combinations can be understood as follows: To select r objects from a set of n objects, we first calculate the number of permutations of r objects taken from n objects using the formula P(n,r). Then we divide by the number of ways to arrange the selected objects using the formula P(r,r). This way we eliminate duplicates and only count distinct selections. For example, to select 3 letters from a set of 5 letters (A, B, C, D, E), we have: $$C(5,3) = \fracP(5;3)P(3;3) = \frac606 = 10$$ The formula for combinations assumes that the objects are distinct and that the order does not matter. If the objects are not distinct or if the order matters, then we need to modify the formula accordingly. For example, if we have a set of 4 letters (A, A, B, C) and we want to select them in different ways without regard to order, we cannot use the formula C(4;4) = 1 because some of the selections are identical (e.g., AA and AA). To avoid counting duplicates, we need to divide by the number of ways to select the indistinguishable objects. In this case, we have two A's that can be selected in C(2;2) ways. Therefore, the number of combinations of 4 letters with two A's is: $$C(4;2) = \fracC(4;4)C(2;2) = \frac11 = 1$$ The notation C(4;2) means that we have 4 objects with two indistinguishable objects. Similarly, if we have a set of 5 letters (A, B, C, D, E) and we want to arrange and select 3 of them with regard to order (e.g., ABC and BAC are considered different), then we cannot use the formula C(5;3) = 10 because some of the arrangements are equivalent. To avoid counting duplicates, we need to multiply by the number of ways to arrange the selected objects. In this case, we have 3 objects that can be arranged in P(3;3) ways. Therefore, the number of combinations of 5 letters taken 3 at a time with regard to order is: $$C(5;3) = C(5;3) \times P(3;3) = 10 \times 6 = 60$$ The notation C(5;3) means that we have 5 objects taken 3 at a time with regard to order. Here are some examples of how to use combinations to solve problems: - How many different ways can you choose 2 letters from the word COMBINATION? Solution: We have 11 letters with two O's and two I's. Therefore, $$C(11;2;2) = \fracC(11;2)C(2;2) = \frac551 = 55$$ - How many different five-card hands can be dealt from a standard deck of 52 cards? Solution: We have 52 cards and we want to select 5 of them without regard to order. Therefore, $$C(52;5) = \frac52!5!(52-5)! = \frac52!5!47! = \frac52 \times 51 \times 50 \times 49 \times 485 \times 4 \times 3 \times 2 \times 1 = 2598960$$ - How many ways can a committee of 4 people be chosen from a group of 10 people? Solution: We have 10 people and we want to select 4 of them without regard to order. Therefore, $$C(10;4) = \frac10!4!(10-4)! = \frac10!4!6! = \frac10 \times 9 \times 8 \times 74 \times 3 \times 2 \times 1 = 210$$ ## Factorials A factorial is the product of all positive integers from 1 to a given number. For example, $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$ Factorials are used to calculate permutations and combinations, as well as other mathematical expressions. By convention, we define $$0! = 1$$ This is because there is only one way to arrange zero objects, which is to do nothing. Some properties and rules of factorials are: - $$n! = n(n-1)!$$ for any positive integer n. For example, $$6! = 6 \times 5! = 6 \times 120 = 720$$ - $$n! > n^n$$ for any positive integer n greater than or equal to 3. For example, $$4! = 24 > 4^4 =16$$ - $$\lim_n\to\infty n! = \infty$$ This means that factorials grow very fast as n increases. - $$\lim_n\to\infty \fracn!n^n =0$$ This means that factorials grow slower than exponential functions as n increases. ## Binomial Coefficients A binomial coefficient is a number that represents the number of combinations of r objects taken from a set of n objects. For example, $$5\choose3 = C(5;3) =\frac5!3!(5-3)!=\frac101=10$$ Binomial coefficients are also written as $_nC_r$ or $C^n_r$. They are called binomial coefficients because they appear in the expansion of binomials (expressions with two terms). For example, $$(x+y)^2=x^2+2xy+y^2=2\choose0x^2+2\choose1xy+2\choose2y^2$$ Some properties and rules of binomial coefficients are: - $$n\choose0=n\choose n=1$$ for any non-negative integer n. This is because there is only one way to select zero or all objects from a set. - objects from a set of n objects is equivalent to selecting n-r objects that are not selected. - $$n\choose r=n-1\choose r-1+n-1\choose r$$ for any positive integers n and r. This is known as Pascal's identity. It can be proved by considering two cases: either the n-th object is selected or not. - $$\sum_r=0^nn\choose r=2^n$$ for any non-negative integer n. This is because there are 2^n possible subsets of a set of n objects. - $$\sum_r=0^n(-1)^rn\choose r=0$$ for any positive integer n. This is because the sum of the terms with even and odd values of r cancel out. ## Pascal's Triangle Pascal's triangle is a triangular array of numbers that contains the binomial coefficients as its entries. It is constructed by starting with 1 at the top and then adding two adjacent numbers to get the next number below them. For example, ![Pascal's triangle](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif) Pascal's triangle is related to binomial coefficients because each row corresponds to the coefficients in the expansion of (x+y)^n, where n is the row number starting from zero. For example, $$(x+y)^3=x^3+3x^2y+3xy^2+y^3=3\choose0x^3+3\choose1x^2y+3\choose2xy^2+3\choose3y^3$$ Pascal's triangle also has many patterns and applications, such as: - The sum of the entries in each row is equal to 2^n, where n is the row number starting from zero. - The sum of the entries in each diagonal is equal to a Fibonacci number, where the Fibonacci sequence is defined by F_0=0, F_1=1, and F_n=F_(n-1)+F_(n-2) for n>1. - The entries in each row are symmetric, meaning that they are the same when read from left to right or from right to left. - The entries in each row are also the coefficients in the expansion of (1+x)^n, where n is the row number starting from zero. For example, $$(1+x)^4=1+4x+6x^2+4x^3+x^4=4\choose0+4\choose1x+4\choose2x^2+4\choose3x^3+4\choose4x^4$$ ## The Binomial Theorem The binomial theorem is a formula that gives the expansion of a binomial raised to any power. It states that: $$(x+y)^n=\sum_r=0^nn\choose rx^n-ry^r$$ for any non-negative integer n and any real numbers x and y. The binomial theorem can be proved by using induction or by using combinatorial arguments. The binomial theorem is related to binomial coefficients because each term in the expansion has a coefficient that is equal to a binomial coefficient. For example, $$(x+y)^5=5\choose0x^5+5\choose1x^4y+5\choose2x^3y^2+5\choose3x^2y^3+5\choose4xy^4+5\choose5y^5$$ The binomial theorem is also related to Pascal's triangle because each row in Pascal's triangle corresponds to the coefficients in the expansion of (x+y)^n, where n is the row number starting from zero. The binomial theorem can be used to expand binomials quickly and easily. For example, $$(2+x)^6=6\choose02^6+6\choose12^5x+6\choose22^4x^2+6\choose32^3x^3+6\choose42^2x^4+6\choose52x^5+6\choose6x^6$$ $$=64+192x+240x^2+160x^3+64x^4+12x^5+x^6$$ ## The Multinomial Theorem The multinomial theorem is a generalization of the binomial theorem that gives the expansion of a sum of more than two terms raised to any power. It states that: $$(x_1+x_2+\cdots+x_k)^n=\sum_r_1+r_2+\cdots+r_k=nn\choose r_1,r_2,\cdots,r_kx_1^r_1x_2^r_2\cdots x_k^r_k$$ for any non-negative integer n and any real numbers x_1, x_2, ..., x_k. The summation is over all non-negative integers r_1, r_2, ..., r_k that add up to n. The coefficient in each term is equal to a multinomial coefficient, which is defined by: $$n\choose r_1,r_2,\cdots,r_k=\fracn!r_1!r_2!\cdots r_k!$$ The multinomial theorem can be proved by using induction or by using combinatorial arguments. The multinomial theorem is a generalization of the binomial theorem because when k=2, it reduces to the binomial theorem. For example, $$(x+y)^4=4\choose0,4x^0y^4+4\choose1,3x^1y^3+4\choose2,2x^2y^2+4\choose3,1x^3y^1+4\choose4,0x^4y^0$$ $$=4\choose04\choose4x^0y^4+4\choose14\choose3x^1y^3+4\choose24\choose2x^2y^2+4\choose34\choose1x^3y^1+4\choose44\choose0x^4y^0$$ $$=4\choose0y^4+4\choose1xy^3+4\choose2x^2y^2+4\choose3x^3y+4\choose4x^4$$ The multinomial theorem can be used to expand sums of more than two terms quickly and easily. For example, $$(1+x+x^2)^3=3\choose0,0,3(1)^0(x)^0(x^2)^3+3\choose0,1,2(1)^0(x)^1(x^2)^2+3\choose0,2,1(1)^0(x)^2(x^2)^1$$ - 2+3\choose1,1,1(1)^1(x)^1(x^2)^1+3\choose1,2,0(1)^1(x)^2(x^2)^0$$ $$+3\choose2,0,1(1)^2(x)^0(x^2)^1+3\choose2,1,0(1)^2(x)^1(x^2)^0+3\choose3,0,0(1)^3(x)^0(x^2)^0$$ $$=x^6+3x^5+6x^4+7x^3+6x^2+3x+1$$ ## Permutations with Indistinguishable Objects Sometimes we have a set of objects that are not all distinct. For example, we may have a set of 10 balls with 4 red balls, 3 blue balls, and 3 green balls. In this case, we cannot use the formula for permutations with distinct objects because some of the arrangements are identical. For example, swapping two red balls does not change the arrangement. To calculate the number of permutations with indistinguishable objects, we need to divide by the number of ways to arrange the indistinguishable objects within each group. For example, $$P(10;4;3;3) = \frac10!4!3!3! = \frac10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 14 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1 = 4200$$ The notation P(10;4;3;3) means that we have 10 objects with 4 indistinguishable objects of one type, 3 indistinguishable objects of another type, and 3 indistinguishable objects of another type. Here are some examples of how to use permutations with indistinguishable objects to solve problems: - How many di