Yes
The solution is count the number of letters in the word and divide by the number of permutations of the repeated letters; 7!/3! = 840.
attendant
The number of permutations of the letters SWIMMING is 8 factorial or 40,320. The number of distinct permutations, however, due to the duplication of the letters I and M is a factor of 4 less than that, or 10,080.
The number of permutations of the letters in the word LOUISIANA is 9 factorial or 362,880. However, since the letters I and A are each repeated once, you need to divide that by 4 to determine the number of distinct permutations, giving you 90,720.
We know there are 10! (ten factorial) permutations (that's about 3,628,800 permutations); however, we know that number includes repeated permutations, as there are 3 s', 3 t's and 2 i's. So we have to divide by the number of ways these can be written as individual permutations (if they were considered as unique elements), which are 3! (= 6), 3! and 2! (= 2) respectively. So our final calculation would be 3628800 / (6 * 6 * 2) = 50400 unique permutations.
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.
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 n!.
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.
Yes
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.
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.
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
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
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).
No, sometimes they will be equal (when all items being permutated are all different, eg all permutations of {1, 2, 3} are distinguishable).
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