Permutation and combination

This article describes the formula syntax and usage of the PERMUT function in Microsoft Excel. Description Returns the number of permutations for a given number of objects that can be selected from number objects. A permutation is any set or subset of objects or events where internal order is significant. Permutations are different from combinations, for which the internal order is not significant. Use this function for lottery-style probability calculations. Syntax PERMUT(number, number_chosen) The PERMUT function syntax has the following arguments: • Number Required. An integer that describes the number of objects. • Number_chosen Required. An integer that describes the number of objects in each permutation. Remarks • Both arguments are truncated to integers. • If number or number_chosen is nonnumeric, PERMUT returns the #VALUE! error value. • If number ≤ 0 or if number_chosen < 0, PERMUT returns the #NUM! error value. • If number < number_chosen, PERMUT returns the #NUM! error value. • The equation for the number of permutations is: Example Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data. Data Description 100 Number of objects 3 Number of objects in each permutation Formula Description R esult =PERMUT(A2,A3) Permutations possible for the arguments specified in A2:A3. 970200 =PERMUT...

Permutations and Combinations Questions Permutations and Combinations questions are provided here, along with detailed explanations to make the students understand easily. Permutation and combinations contain a large number of applications in our daily life. Thus, it is essential to learn and practise the fundamentals of these concepts. Also, we know that permutation and combination is one of the important chapters of Class 11 Maths. In this article, you will find solved questions and practice questions on permutations and combinations. However, these questions cover easy, medium and high difficulty levels. What are permutations and combinations? A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Also, read: Permutations and Combinations Questions and Answers 1. How many numbers are there between 99 and 1000, having at least one of their digits 7? Solution: Numbers between 99 and 1000 are all three-digit numbers. Total number of 3 digit numbers having at least one of their digits as 7 = (Total numbers of three-digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all) = (9 × 10 × 10) – (8 × 9 × 9) = 900 – 648 = 252 2. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more t...

You're talking about a permutation, even though in the real world people use the word combination (which is mathematically wrong). Here's an easy way to remember: - If a group consisting of Alice, Bob and Charlie has the same meaning as a group consisting of Charlie, Alice and Bob, you're talking about a combination. - If a group consisting of Alice, then Bob and THEN Charlie is NOT the same meaning as a group consisting of Charlie, then Alice and THEN Bob, you're talking about a permutation. Remember to upvote good questions and helpful answers Are permutations and combinations the same thing? I thought that in combinations you couldn't use the same people. Like in combinations we could do :CBA but we couldn't do BCA because they are the same people. And in permutations we could do :CBA and BCA because order didn't matter. I'm soooo confused! In Permutations the order matters. So ABC would be one permutation and ACB would be another, for example. In Combinations ABC is the same as ACB because you are combining the same letters (or people). Now, there are 6 (3 factorial) permutations of ABC. Therefore, to calculate the number of combinations of 3 people (or letters) from a set of six, you need to divide 6! by 3!. I think its best to write out the combinations and permutations like Sal does; that really helps me out. 4:50 Salman says, to find the number of ways to arrange three people from the six, the equation is 120 (total number of permutations) divided by 6 (number of w...

\( \newcommand^n m_i\). This too can be proved by induction. Example How many outcomes are possible when three dice are rolled, if no two of them may be the same? The first two dice together have \(6\cdot 5=30 \) possible outcomes, from above. For each of these 30 outcomes, there are four possible outcomes for the third die, so the total number of outcomes is \(30\cdot 4=6\cdot 5\cdot 4=120\). (Note that we consider the dice to be distinguishable, that is, a roll of 6, 4, 1 is different than 4, 6, 1, because the first and second dice are different in the two rolls, even though the numbers as a set are the same.) Example \(\PageIndex\) Suppose blocks numbered 1 through \(n \) are in a barrel; we pull out \(k \) of them, placing them in a line as we do. How many outcomes are possible? That is, how many different arrangements of \(k \) blocks might we see? This is essentially the same as the previous example: there are \(k \) "spots'' to be filled by blocks. Any of the \(n \) blocks might appear first in the line; then any of the remaining \(n-1 \) might appear next, and so on. The number of outcomes is thus \(n(n-1)(n-2)\cdots(n-k+1)\), by the multiplication principle. In the previous example, the first "spot'' was die number one, the second spot was die number two, the third spot die number three, and \(6\cdot5\cdot4=6(6-1)(6-2)\); notice that \(6-2=6-3+1\). This is quite a general sort of problem: Definition \(\PageIndex.\nonumber\]This is correct only if \(0!=1\), so we a...

\( \newcommand\) • • • • Learning Objectives • Calculate the probability of two independent events occurring • Define permutations and combinations • List all permutations and combinations • Apply formulas for permutations and combinations This section covers basic formulas for determining the number of various possible types of outcomes. The topics covered are: • counting the number of possible orders • counting using the multiplication rule • counting the number of permutations • counting the number of combinations Possible Orders Suppose you had a plate with three pieces of candy on it: one green, one yellow, and one red. You are going to pick up these three pieces one at a time. The question is: In how many different orders can you pick up the pieces? Table \(\PageIndex \] This means that if there were \(5\) pieces of candy to be picked up, they could be picked up in any of \(5! = 120\) orders. Multiplication Rule Imagine a small restaurant whose menu has \(3\) soups, \(6\) entrées, and \(4\) desserts. How many possible meals are there? The answer is calculated by multiplying the numbers to get \(3 \times 6 \times 4 = 72\). You can think of it as first there is a choice among \(3\) soups. Then, for each of these choices there is a choice among \(6\) entrées resulting in \(3 \times 6 = 18\) possibilities. Then, for each of these \(18\) possibilities there are \(4\) possible desserts yielding \(18 \times 4 = 72\) total possibilities. Permutations Suppose that there were ...

This article describes the formula syntax and usage of the PERMUT function in Microsoft Excel. Description Returns the number of permutations for a given number of objects that can be selected from number objects. A permutation is any set or subset of objects or events where internal order is significant. Permutations are different from combinations, for which the internal order is not significant. Use this function for lottery-style probability calculations. Syntax PERMUT(number, number_chosen) The PERMUT function syntax has the following arguments: • Number Required. An integer that describes the number of objects. • Number_chosen Required. An integer that describes the number of objects in each permutation. Remarks • Both arguments are truncated to integers. • If number or number_chosen is nonnumeric, PERMUT returns the #VALUE! error value. • If number ≤ 0 or if number_chosen < 0, PERMUT returns the #NUM! error value. • If number < number_chosen, PERMUT returns the #NUM! error value. • The equation for the number of permutations is: Example Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data. Data Description 100 Number of objects 3 Number of objects in each permutation Formula Description R esult =PERMUT(A2,A3) Permutations possible for the arguments specified in A2:A3. 970200 =PERMUT...

You're talking about a permutation, even though in the real world people use the word combination (which is mathematically wrong). Here's an easy way to remember: - If a group consisting of Alice, Bob and Charlie has the same meaning as a group consisting of Charlie, Alice and Bob, you're talking about a combination. - If a group consisting of Alice, then Bob and THEN Charlie is NOT the same meaning as a group consisting of Charlie, then Alice and THEN Bob, you're talking about a permutation. Remember to upvote good questions and helpful answers Are permutations and combinations the same thing? I thought that in combinations you couldn't use the same people. Like in combinations we could do :CBA but we couldn't do BCA because they are the same people. And in permutations we could do :CBA and BCA because order didn't matter. I'm soooo confused! In Permutations the order matters. So ABC would be one permutation and ACB would be another, for example. In Combinations ABC is the same as ACB because you are combining the same letters (or people). Now, there are 6 (3 factorial) permutations of ABC. Therefore, to calculate the number of combinations of 3 people (or letters) from a set of six, you need to divide 6! by 3!. I think its best to write out the combinations and permutations like Sal does; that really helps me out. 4:50 Salman says, to find the number of ways to arrange three people from the six, the equation is 120 (total number of permutations) divided by 6 (number of w...

\( \newcommand^n m_i\). This too can be proved by induction. Example How many outcomes are possible when three dice are rolled, if no two of them may be the same? The first two dice together have \(6\cdot 5=30 \) possible outcomes, from above. For each of these 30 outcomes, there are four possible outcomes for the third die, so the total number of outcomes is \(30\cdot 4=6\cdot 5\cdot 4=120\). (Note that we consider the dice to be distinguishable, that is, a roll of 6, 4, 1 is different than 4, 6, 1, because the first and second dice are different in the two rolls, even though the numbers as a set are the same.) Example \(\PageIndex\) Suppose blocks numbered 1 through \(n \) are in a barrel; we pull out \(k \) of them, placing them in a line as we do. How many outcomes are possible? That is, how many different arrangements of \(k \) blocks might we see? This is essentially the same as the previous example: there are \(k \) "spots'' to be filled by blocks. Any of the \(n \) blocks might appear first in the line; then any of the remaining \(n-1 \) might appear next, and so on. The number of outcomes is thus \(n(n-1)(n-2)\cdots(n-k+1)\), by the multiplication principle. In the previous example, the first "spot'' was die number one, the second spot was die number two, the third spot die number three, and \(6\cdot5\cdot4=6(6-1)(6-2)\); notice that \(6-2=6-3+1\). This is quite a general sort of problem: Definition \(\PageIndex.\nonumber\]This is correct only if \(0!=1\), so we a...

Permutations and Combinations Questions Permutations and Combinations questions are provided here, along with detailed explanations to make the students understand easily. Permutation and combinations contain a large number of applications in our daily life. Thus, it is essential to learn and practise the fundamentals of these concepts. Also, we know that permutation and combination is one of the important chapters of Class 11 Maths. In this article, you will find solved questions and practice questions on permutations and combinations. However, these questions cover easy, medium and high difficulty levels. What are permutations and combinations? A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Also, read: Permutations and Combinations Questions and Answers 1. How many numbers are there between 99 and 1000, having at least one of their digits 7? Solution: Numbers between 99 and 1000 are all three-digit numbers. Total number of 3 digit numbers having at least one of their digits as 7 = (Total numbers of three-digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all) = (9 × 10 × 10) – (8 × 9 × 9) = 900 – 648 = 252 2. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more t...

\( \newcommand\) • • • • Learning Objectives • Calculate the probability of two independent events occurring • Define permutations and combinations • List all permutations and combinations • Apply formulas for permutations and combinations This section covers basic formulas for determining the number of various possible types of outcomes. The topics covered are: • counting the number of possible orders • counting using the multiplication rule • counting the number of permutations • counting the number of combinations Possible Orders Suppose you had a plate with three pieces of candy on it: one green, one yellow, and one red. You are going to pick up these three pieces one at a time. The question is: In how many different orders can you pick up the pieces? Table \(\PageIndex \] This means that if there were \(5\) pieces of candy to be picked up, they could be picked up in any of \(5! = 120\) orders. Multiplication Rule Imagine a small restaurant whose menu has \(3\) soups, \(6\) entrées, and \(4\) desserts. How many possible meals are there? The answer is calculated by multiplying the numbers to get \(3 \times 6 \times 4 = 72\). You can think of it as first there is a choice among \(3\) soups. Then, for each of these choices there is a choice among \(6\) entrées resulting in \(3 \times 6 = 18\) possibilities. Then, for each of these \(18\) possibilities there are \(4\) possible desserts yielding \(18 \times 4 = 72\) total possibilities. Permutations Suppose that there were ...