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∙ 11y agoThere 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.
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∙ 11y agothree
7 factorial
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.
act
three
7 factorial
120?
2520.
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.
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
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 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.
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.
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.