Letter S
Six.
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
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."
English is spoken by the largest number of countries. It is an official language in 59 countries.
The number of 5 letter arrangements of the letters in the word DANNY is the same as the number of permutations of 5 things taken 5 at a time, which is 120. However, since the letter N is repeated once, the number of distinct permutations is one half of that, or 60.
The number of permutations of the letter ABCDEF is 6 factorial, or 720.
The number of permutations of the letters EFFECTIVE is 9 factorial or 362,880. Since the letter E is repeated twice we need to divide that by 4, to get 90,720. Since the letter F is repeated once we need to divide that by 2, to get 45,360.
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
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 number of permutations of the letters MASS where S needs to be the first letter is the same as the number of permutations of the letters MAS, which is 3 factorial, or 6. SMAS SMSA SAMS SASM SSMA SSAM
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.