No, sometimes they will be equal (when all items being permutated are all different, eg all permutations of {1, 2, 3} are distinguishable).
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.
The letters from a to J consist of 10 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 10!. Therefore, the total number of permutations is 10! = 3,628,800.
Since there are no duplicate letters in the word RAINBOW, the number of permutations of those letters is simply the number of permutations of 7 things taken 7 at a time, i.e. 7 factorial, which is 5040.
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.
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.
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!.
The letters from a to J consist of 10 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 10!. Therefore, the total number of permutations is 10! = 3,628,800.
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
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.
not sure
attendant
Since the word MATH does not have any duplicated letters, the number of permutations of those letters is simply the number of permutations of 4 things taken 4 at a time, or 4 factorial, or 24.
Any 6 from 12 = 924 permutations
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.
Since there are no duplicate letters in the word RAINBOW, the number of permutations of those letters is simply the number of permutations of 7 things taken 7 at a time, i.e. 7 factorial, which is 5040.