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For the first letter, you can choose any of the four.
For the second letter, you can choose any of the remaining three.
For the third letter, you can choose either of the remaining two.
For the fourth letter, you don't get a choice, you have to use the one that's left.
So 4x3x2x1 = 24 possible permutations.
What else can I help you with?
How many permutation are possible of the letters in the word numbers?
The word "numbers" consists of 7 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 7!. Therefore, the total number of permutations is 7! = 5,040.
How many permutations are possible of the word rainbow?
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.
How many permutations of the letters in the word noon are there?
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."
How many distinguishable and indistinguishable permutations are there in the word algrebra?
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.
How many permutations are in the word freezer?
The word "freezer" has 7 letters, with the letter "e" appearing twice and the letter "r" appearing twice. The number of distinct permutations can be calculated using the formula for permutations of a multiset: ( \frac{n!}{n_1! , n_2! , \ldots , n_k!} ), where ( n ) is the total number of letters and ( n_i ) are the frequencies of the repeated letters. Thus, the number of permutations is ( \frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260 ). Therefore, there are 1,260 distinct permutations of the word "freezer."
How many permutations are in the word math?
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.
What is the number of distinguishable permutations of the letters in the word GLASSES?
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.
How many permutations of the letters in the word Louisiana are there?
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.
How many permutations are in the word united?
UNITED = 6 letters The letters in the word UNITED did not repeat so the number of permutations = 6! = 6x5x4x3x2 =720
How many permutation are possible of the letters in the word numbers?
The word "numbers" consists of 7 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 7!. Therefore, the total number of permutations is 7! = 5,040.
How many arrangements of the letters in the word SCHOOLS are there?
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.
How many permutations are possible of the word rainbow?
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.
How many permutations of letters in word swimming?
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.
What is the number of distinguishable permutations of the letters in the word oregon?
360. There are 6 letters, so there are 6! (=720) different permutations of 6 letters. However, since the two 'o's are indistinguishable, it is necessary to divide the total number of permutations by the number of permutations of the letter 'o's - 2! = 2 Thus 6! ÷ 2! = 360
How many distinguishable and indistinguishable permutations are there in the word algrebra?
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.
How many permutations are in the word mathematics?
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.
How many permutations are there of he letters in the word greet?
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).
