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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.
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.
We can clearly observe that the word "ellises" has 7 letters and three pairs of letters are getting repeated that are 'e','l' and 's'. So, Number of distinguishable permutations = 7!/(2!2!2!) = 7 x 6 x 5 x 3 = 630.
You have six different letters. They can be arranged in 6 x 5 x 4 x 3 x 2 = 720 different ways.
three
120?
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.
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 solution is count the number of letters in the word and divide by the number of permutations of the repeated letters; 7!/3! = 840.
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.
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.
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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
Normally, there would be 5!=120 different permutations* of five letters. Since two of the letters are the same, we can each of these permutations will be duplicated once (with the matching letters switched). So there are only half as many, or 60 permutations.* (the correct terminology is "permutation". "combination" means something else.)
We can clearly observe that the word "ellises" has 7 letters and three pairs of letters are getting repeated that are 'e','l' and 's'. So, Number of distinguishable permutations = 7!/(2!2!2!) = 7 x 6 x 5 x 3 = 630.
2520.