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Q: Is the number of permutations of 2 items from the same data set always 2 times the number of combiations when taking 2 objects from a data set?
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Does the number of permutations always exceed the number of combinations?

No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.


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)!


The number of permutations of n objects taken all together is?

The number of permutations of n objects taken all together is n!.


How can you find the number of permutations of a set of objects?

If there are n different objects, the number of permutations is factorial n which is also written as n! and is equal to 1*2*3*...*(n-1)*n.


Is the number of permutations of two items from a data set is always two times the number of combinations when taking two objects at a time from the same data set?

Yes


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 do you do permutations with repeating symbols?

Suppose you have n objects and within those, there arem1 objects of kind 1m2 objects of kind 2and so on.Then the number of permutations of the n objects is n!/[m1!* m2!...]For example, permutations of the word "banana"n = 6there are 3 "a"s so m1 = 3there are 2 "n"s so m2 = 2therefore, the number of permutations = 6!/(3!*2!) = 720/(3*2) = 120.


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 do you calculate distinguishable permutations?

The number of permutations of n distinct objects is n! = 1*2*3* ... *n. If a set contains n objects, but k of them are identical (non-distinguishable), then the number of distinct permutations is n!/k!. If the n objects contains j of them of one type, k of another, then there are n!/(j!*k!). The above pattern can be extended. For example, to calculate the number of distinct permutations of the letters of "statistics": Total number of letters: 10 Number of s: 3 Number of t: 3 Number of i: 2 So the answer is 10!/(3!*3!*2!) = 50400


What is the permutation of 5?

Permutations are the different arrangements of any number of objects. When we arrange some objects in different orders, we obtain different permutations.Therefore, you can't say "What is the permutation of 5?". To calculate permutations, one has to get the following details:The total number of objects (n) (necessary)The number of objects taken at a time (r) (necessary)Any special conditions mentioned in the question (optional).


Will the number of permutations alwaysbe greater than the number of distinguisible permutations?

No, sometimes they will be equal (when all items being permutated are all different, eg all permutations of {1, 2, 3} are distinguishable).


What is the number of distinguishable permutations of the letters in the word oregon?

360. There are 6 letters, so there are 6! (=720) different permutations of 6 letters. However, since the two 'o's are indistinguishable, it is necessary to divide the total number of permutations by the number of permutations of the letter 'o's - 2! = 2 Thus 6! ÷ 2! = 360