You have six different letters. They can be arranged in 6 x 5 x 4 x 3 x 2 = 720 different ways.
three
7 factorial
There are 7 factorial, or 5,040 permutations of the letters of ALGEBRA. However, only 2,520 of them are distinguishable because of the duplicate A's.
There are 7 factorial, or 5,040 permutations of the letters of OCTOBER. However, only 2,520 of them are distinguishable because of the duplicate O's.
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.
act
three
7 factorial
120?
There are 7 factorial, or 5,040 permutations of the letters of ALGEBRA. However, only 2,520 of them are distinguishable because of the duplicate A's.
2520.
The word mathematics has 11 letters; 2 are m, a, t. The number of distinguishable permutations is 11!/(2!2!2!) = 39916800/8 = 4989600.
There are 7 factorial, or 5,040 permutations of the letters of OCTOBER. However, only 2,520 of them are distinguishable because of the duplicate O's.
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
The word "rectangle" consists of 9 letters, with the letter 'e' appearing twice and all other letters being unique. To find the number of distinguishable permutations, we use the formula for permutations of a multiset: (\frac{n!}{n_1! \cdot n_2! \cdots n_k!}), where (n) is the total number of letters and (n_i) are the frequencies of the distinct letters. Thus, the number of distinguishable permutations is (\frac{9!}{2!} = \frac{362880}{2} = 181440).
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.