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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 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 MATHEMATICS has 11 letters. The number of permutations of 11 things taken 11 at a time is 11 factorial (11!), or 39,916,800.
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.
We know there are 10! (ten factorial) permutations (that's about 3,628,800 permutations); however, we know that number includes repeated permutations, as there are 3 s', 3 t's and 2 i's. So we have to divide by the number of ways these can be written as individual permutations (if they were considered as unique elements), which are 3! (= 6), 3! and 2! (= 2) respectively. So our final calculation would be 3628800 / (6 * 6 * 2) = 50400 unique permutations.
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."
The word "away" has 4 letters, with the letter "a" repeating twice. To find the number of unique permutations, use the formula for permutations of multiset: ( \frac{n!}{n_1! \cdot n_2! \cdots} ), 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 unique permutations is ( \frac{4!}{2!} = \frac{24}{2} = 12 ).
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.
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.
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 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.
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 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 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