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Total permutations 10! ie factorial 10 = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3628800. The 2 "c"s are interchangeable which halves this figure to 1814400, similarly the "a"s and "l"s are interchangeable which reduces by half twice more, ie to 907200 and then to 453600.

Q: Find the number of distinguishable permutations of the letters in the word calculator?

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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

Take the total number of letters factorial, then divide by the multiple letters factorial (a and e). 7! / (2!*2!) or 1260.

September has 9 letter, of which one appears 3 times. So the number of distinct permutations is 9!/3! = 120,960

There are ten letters in the word. The total number of possible permutations is(10) x (9) x (8) x (7) x (6) x (5) x (4) x (3) x (2) = 3,628,800But the two 'c's can be arranged in either of 2 ways with no distinguishable change.Also, the three 'i's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.And the three 't's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.So the total number of possible permutations can be divided by (2 x 6 x 6) = 72, the number oftimes each distinguishable permutation occurs with different and indisnguishable arrangementsof 'c', 'i', and 't'.We're left with(10) x (9) x (8) x (7) x (...) x (5) x (...) x (...) x (2) = (3,628,800/72) = 50,400 distinguishable arrangements.

If there are n objects and you have to choose r objects then the number of permutations is (n!)/((n-r)!). For circular permutations if you have n objects then the number of circular permutations is (n-1)!

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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.

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 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 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.

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.

It is 6! = 6*5*4*3*2*1 = 720

Take the total number of letters factorial, then divide by the multiple letters factorial (a and e). 7! / (2!*2!) or 1260.

2520.

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 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 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