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."
There are 7 factorial, or 5040 permutations of the letters in the word NUMBERS.
three
7 factorial
239,500,800 12!/2! * * * * * Actually, as the word "permutation" [not permutations] has 11 letters, the answer is 11!/2! = 19,958,400
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.
There are 6! = 720 permutations.
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
There are 7 factorial, or 5040 permutations of the letters in the word NUMBERS.
There are 9,979,200
quiet
tt
UNITED = 6 letters The letters in the word UNITED did not repeat so the number of permutations = 6! = 6x5x4x3x2 =720
2520
three
7 factorial
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.
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.