0
The number of permutations of r objects taken from n distinct objects is nPr = n!/(n-r)!
n! = 1*2*3*... *(n-1)*n
In many cases, one can do permutation problems without the "formula."
Here is a concrete example, then a more abstract version to help see this.
The purpose of pointing this out is not only to make it easier to do this type of problem, but to help understand and remember the formula.
Say we have 10 distinct (we assume distinct from now on) objects and we want to pick 3 objects from the 10. The order of choice is important.
Since there are 10 objects, there are 10 choices when we pick the first one. Now there are 9 left since one is picked so there are 9 choices for the next object. Similarly, there are 8 choices for the third object. The multiplication rule tells us there are 10x9x8=720 ways to pick these three objects.
Now say we have n distinct objects and we want to pick r of them. We have n choices for the first, n-1 choices for the second and n-3 choices for the third.
If n=10 as above and r=3, we saw this was 10x9x8 but in the more general case it is n x (n-1) x(n-2). But how do we know where to stop? We want to pick r items.
So in the case of r=3, the last number in our product was 8. This is (10-3)+1. So in general the last number in the product is (n-r)+1
The general pattern therefore for picking r objects taken from n distinct objects is
n(n-1)(n-2)(n-3)...(n-r+1). This is the same as n!/(n-r)!
The denominator in the formula simply allows us to write n! which is 1*2*3*... *(n-1)*n
and cancel out all the numbers in the product that are less than n-r+1. For example. Using the same example if we pick 3 items from 10 we saw that was 10x9x8. Using the formula we have 10!/(10-3)! or 10!/7! which is (10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1) and after cancelling the (7x6x5x4x3x2x1), which occurs in the products in both the numerator and the denominator, we are left with 10x9x8=720 which we know to be the answer.
What else can I help you with?
What is a general formula for the permutation nPr in terms of n and r For instance if you have the permutation nP1 how do you find n?
Permutation Formula A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk . Formula:Example:How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15. different lineups
How many permutations in obfuscation?
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
How many permutations are possible of the word information?
39916800 permutations are possible for the word INFORMATION.
How do you find permutation?
If there are n objects and you have to choose r objects then the number of permutations is (n!)/((n-r)!). For circular permutations if you have n objects then the number of circular permutations is (n-1)!
What is the correct number of permutations of 6 items from a list of 12 items?
Any 6 from 12 = 924 permutations
What is the formula to find out how many words you can make with a word?
There is no such formula since most of the possible permutations will not be words.
What is the number of distinguishable permutations?
The formula for finding the number of distinguishable permutations is: N! -------------------- (n1!)(n2!)...(nk!) where N is the amount of objects, k of which are unique.
How many distinguishable and indistinguishable permutations are there in the word algrebra?
The word "algrebra" has 8 letters, with the letter 'a' appearing twice and 'r' appearing twice. To find the number of distinguishable permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2!} ), where ( n ) is the total number of letters and ( n_1, n_2 ) are the frequencies of the repeating letters. Thus, the number of distinguishable permutations is ( \frac{8!}{2! \times 2!} = 10080 ). Since all letters are counted in this formula, there are no indistinguishable permutations in this context.
What is the formula for finding permutations?
The number of permutations of r objects selected from n different objects is nPr = n!/(n-r)! where n! denotes 1*2*3*,,,*n and also, 0! = 1
How many permutations are in the word freezer?
The word "freezer" has 7 letters, with the letter "e" appearing twice and the letter "r" appearing twice. The number of distinct permutations can be calculated using the formula for permutations of a multiset: ( \frac{n!}{n_1! , n_2! , \ldots , n_k!} ), where ( n ) is the total number of letters and ( n_i ) are the frequencies of the repeated letters. Thus, the number of permutations is ( \frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260 ). Therefore, there are 1,260 distinct permutations of the word "freezer."
How many permutations of the letters in the word noon are there?
The word "noon" consists of 4 letters, where 'n' appears twice and 'o' appears twice. To find the number of distinct permutations, we use the formula for permutations of multiset: ( \frac{n!}{n_1! \cdot n_2!} ), where ( n ) is the total number of letters and ( n_1, n_2 ) are the frequencies of the repeating letters. Thus, the number of permutations is ( \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6 ). Therefore, there are 6 distinct permutations of the letters in the word "noon."
How many different permutations are there of 4 objects taken 4 at a time?
The number of different permutations of 4 objects taken 4 at a time is calculated using the formula ( n! ), where ( n ) is the number of objects. For 4 objects, this is ( 4! = 4 \times 3 \times 2 \times 1 = 24 ). Therefore, there are 24 different permutations.
What is a general formula for the permutation nPr in terms of n and r For instance if you have the permutation nP1 how do you find n?
Permutation Formula A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk . Formula:Example:How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15. different lineups
How many permutations does the string ABC have?
3x2x1=6 permutations.
How many permutations can you get out of an eight word phrase?
There are 8! = 40320 permutations.
How many permutations are there of he letters in the word greet?
The word "greet" consists of 5 letters, where 'g', 'r', and 't' are unique, and 'e' appears twice. To find the number of distinct permutations, we use the formula for permutations of multiset: (\frac{n!}{n_1! \cdot n_2! \cdot \ldots}), where (n) is the total number of letters and (n_1, n_2, \ldots) are the frequencies of the repeated letters. Thus, the number of permutations is (\frac{5!}{2!} = \frac{120}{2} = 60).
What is the permutations and combinations for the word great?
There are 120 permutations and only 1 combination.
