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 "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."
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 120 permutations and only 1 combination.
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."
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 120 permutations and only 1 combination.
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
There are 6! = 720 permutations.
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.