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135: including 15 with two Os which are assumed to be indistinguishable.

If one of the Os is upper case and the other lower (so that Oct is different from oct), then the answer is 8*7*6 = 336.

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Q: How many 3 letter permutations of October are there?
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How many 3 letter permutations of the word October?

There are 195 3-letter permutations.


How many different 3 letter permutations can be formed from the letters in the word count?

There are 5*4*3 = 60 permutations.


How many permutations exist in a four letter word?

4! 4 * 3 * 2 * 1 = 24


How many permutations are there in any 3 from 5?

There are 5P3 = 5!/2! = 5*4*3 = 60 permutations.


How many permutations are in the word peak?

The first letter of the perm can be any of the 4, the second any of the remaining 3, the third either of the remaining 2. There are thus 4 x 3 x 2 ie 24 permutations.


How many 3 letter arrangements are possible for the word bookkeeper?

Permutations of 10 letters taken 3 a time = 10 x 9 x 8 = 720


How many possible permutations using the word bite?

4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------


How many permutations of 3 different digits are there chosen from the ten digits 0 to 9 inclusive?

How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?


How many different ways are there to to scramble to letters of the word mass if s needs to be the first letter?

The number of permutations of the letters MASS where S needs to be the first letter is the same as the number of permutations of the letters MAS, which is 3 factorial, or 6. SMAS SMSA SAMS SASM SSMA SSAM


How many distinct permutations are there of the word statistics?

We know there are 10! (ten factorial) permutations (that's about 3,628,800 permutations); however, we know that number includes repeated permutations, as there are 3 s', 3 t's and 2 i's. So we have to divide by the number of ways these can be written as individual permutations (if they were considered as unique elements), which are 3! (= 6), 3! and 2! (= 2) respectively. So our final calculation would be 3628800 / (6 * 6 * 2) = 50400 unique permutations.


How many permutations are there in a 3-letter code which can use the letters from A to Z inclusive?

A - Z means you can use the whole alphabet, which usually contains 26 letters. So a one-letter code would give you 26 permutations. 2 letters will give you 26 x 26 permutations. A three letter code, finally, will give you 26 x 26 x 26 , provided you don't have any restrictions given, like avoiding codes formed from 3 similar letters and such.


How many permutations are in the word stop?

There are 4*3*2*1 = 24 of them.