The number of permutations of n distinct objects is n! = 1*2*3* ... *n.
If a set contains n objects, but k of them are identical (non-distinguishable), then the number of distinct permutations is n!/k!.
If the n objects contains j of them of one type, k of another, then there are n!/(j!*k!).
The above pattern can be extended. For example, to calculate the number of distinct permutations of the letters of "statistics":
Total number of letters: 10
Number of s: 3
Number of t: 3
Number of i: 2
So the answer is 10!/(3!*3!*2!) = 50400
The formula for finding the number of distinguishable permutations is: N! -------------------- (n1!)(n2!)...(nk!) where N is the amount of objects, k of which are unique.
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 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.
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. 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.
The formula for finding the number of distinguishable permutations is: N! -------------------- (n1!)(n2!)...(nk!) where N is the amount of objects, k of which are unique.
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.
2520.
cat
act
three
7 factorial
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
120?
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.
What are the different counting techniques