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.
three
7 factorial
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 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).
three
act
7 factorial
120?
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 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 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 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 EFFECTIVE is 9 factorial or 362,880. To determine the distinct permutations, you have to compensate for the three E's (divide by 4) and the two F's (divide by 2), giving you 45,360.