2520.
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 word mathematics has 11 letters; 2 are m, a, t. The number of distinguishable permutations is 11!/(2!2!2!) = 39916800/8 = 4989600.
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
act
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.
120?
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 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.
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.
September has 9 letter, of which one appears 3 times. So the number of distinct permutations is 9!/3! = 120,960