Suppose you have n objects and within those, there arem1 objects of kind 1
m2 objects of kind 2
and so on.
Then the number of permutations of the n objects is n!/[m1!* m2!...]
For example, permutations of the word "banana"
n = 6
there are 3 "a"s so m1 = 3
there are 2 "n"s so m2 = 2
therefore, the number of permutations = 6!/(3!*2!) = 720/(3*2) = 120.
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.
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."
10 x 9 x 8 = 720 different "permutations"
3x2x1=6 permutations.
There are 120 permutations and only 1 combination.
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.
10 x 9 x 8 = 720 different "permutations"
3x2x1=6 permutations.
There are 8! = 40320 permutations.
There are 6! = 720 permutations.
There are 120 permutations and only 1 combination.
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
39916800 permutations are possible for the word INFORMATION.
10! permutations of the word "Arithmetic" may be made.
There are 195 3-letter permutations.
A freezer is a freezer: it has a fixed shape and a fixed purpose so there are really no permutations.
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)!