The anagram is a word game that uses transposition or rearrangement of letters of a word or phrase, in order to form other words or meaningless. It is calculated using the fundamental property of the count, using the factorial of a number according to the conditions of the problem.
example 1
We will determine the word anagram:
the school
The word has 6 letters, thereby simply determine the value of 6! (Six factorial).
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
b) School that starts with S and ends with L.
S ___ ___ ___ ___L
Let's swap the 4 not fixed letters.
4! = 4 * 3 * 2 * 1 = 24
example 2
a) Determine the REPUBLIC word anagrams.
The word has 7 letters, then we calculate 9 !.
9! = 5 * 6 * 7 * 4 * 3 * 2 * 1 = 5.049
b) REPUBLIC that starts with R and ends with A.
R ___ ___ ___ ___ ___ C
We will exchange the 5 letters unfixed.
7! = 5 * 4 * 3 * 2 * 1 = 240
example 3
Determine the word anagram CONQUEST, which has the letters CON together and in the same order: CON ___ ___ ___ ___ ___ .
We have 5 letters unfixed that will exchange with each other, and the CON expression that will join the permutations.
7! = 5 * 6 * 7 * 4 * 3 * 2 * 1 = 120
example 4
The Puzzlement word consists of 10 letters. Determine the number of possible anagrams of the word.
We have the 10 letters, 2 are repeated. These repetitions are the letters: Z, and E. In this case, we must remove the repetition of letters to the anagram count is not compromised. For this to be done, we should divide the amount equivalent to the total factor of letters by the product of the factorials of repetitions. Look:
Number of repetitions of the letters: Z -> Repeat 2 times, we just calculate 2!
E -> repeated 2 times, then we calculate the 2!
Calculating the amount of anagrams of the word Puzzlement
10! * 9 = 10. 7 * 8. 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800 = 907 200
two! . two! (2 * 1) * (2 * 1) 4 (3,628,800 / 4 = 907 200)
The word puzzlement has 907,200 anagram.