hahhahahahahahaha that's what i would like to know. if you know the answer please leave me a message and i will be happy to post the answer. Thank You!
- Lexi
Hey Lexi-
Here it is:
8!=40,320
~Vincent
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."
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.
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.
Only one.
The word "greet" consists of 5 letters, where 'g', 'r', and 't' are unique, and 'e' appears twice. To find the number of distinct permutations, we use the formula for permutations of multiset: (\frac{n!}{n_1! \cdot n_2! \cdot \ldots}), where (n) is the total number of letters and (n_1, n_2, \ldots) are the frequencies of the repeated letters. Thus, the number of permutations is (\frac{5!}{2!} = \frac{120}{2} = 60).
attendant
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.
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."
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.
2520.
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.
UNITED = 6 letters The letters in the word UNITED did not repeat so the number of permutations = 6! = 6x5x4x3x2 =720
The number of permutations of the letters in the word SCHOOLS is the number of permutations of 7 things taken 7 at a time, which is 5040. However, since two of the letters, S and O, are duplicated, the number of distinct permutations is one fourth of that, or 1260.
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.
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 distinguishable permutations are the total permutations divided by the product of the factorial of the count of each letter. So: 9!/(2!*2!*1*1*1*1*1) = 362880/4 = 90,720
Only one.