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.
There can be only one permutation of a single number: so the answer is 7.
To determine the probability of randomly selecting the permutation "abacus" from the letters AABCSU, we first calculate the total number of distinct permutations of the letters. The total permutations are given by the formula ( \frac{n!}{n_1! \cdot n_2! \cdots n_k!} ), where ( n ) is the total number of letters and ( n_i ) are the counts of each distinct letter. Here, we have 6 letters (2 A's, 1 B, 1 C, 1 S, 1 U), resulting in ( \frac{6!}{2!} = 360 ) distinct permutations. Since "abacus" is one specific permutation, the probability is ( \frac{1}{360} ).
nPrwhere:n number of objectsr is number of arrangements
January March April May September October
September has 9 letter, of which one appears 3 times. So the number of distinct permutations is 9!/3! = 120,960
permutation without replacement
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.
There can be only one permutation of a single number: so the answer is 7.
To determine the probability of randomly selecting the permutation "abacus" from the letters AABCSU, we first calculate the total number of distinct permutations of the letters. The total permutations are given by the formula ( \frac{n!}{n_1! \cdot n_2! \cdots n_k!} ), where ( n ) is the total number of letters and ( n_i ) are the counts of each distinct letter. Here, we have 6 letters (2 A's, 1 B, 1 C, 1 S, 1 U), resulting in ( \frac{6!}{2!} = 360 ) distinct permutations. Since "abacus" is one specific permutation, the probability is ( \frac{1}{360} ).
For the first letter, you can have any of 5 choices. Then for the next, you can have any of four. Then for the next, three and so on. Thus the number of permutations can be calculated by 5x4x3x2x1. Doing this gives 120. Therefore the number of permutations of the letters in the word smart is 120.
nPrwhere:n number of objectsr is number of arrangements
The number of 7 letter permutations of the word ALGEBRA is the same as the number of permutation of 7 things taken 7 at a time, which is 5040. However, since the letter A is duplicated once, you have to divide by 2 in order to find out the number of distinct permutations, which is 2520.
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You cannot (answer dated 10 September 2011).
Carbon- sixth element, contains six letters. Mars- fourth planet, contains four letters. Four- fourth number, contains four letters. Seven- seventh number, contains seven letters. September- ninth month, contains nine letters. Saturn- sixth planet, contains six letters. Fifth- fifth number, contains five letters.
January March April May September October