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There are 5*4*3 = 60 permutations.
There are ten letters BUT there are 3 Ss and Ts and 2 Is. So the answer is 10!/(3!*3!*2!) = 50400
420.
This is a basic permutation problem. You have 6 elements, of which you need to select 3, without replacement, and order does matter (123 is different from 321). Permutations are covered by the wikipedia page on permutations. In short, however the answer is P(6, 3) = 6!/(6-3)! = 6*5*4 = 120.
240
There are 5*4*3 = 60 permutations.
There are 8P5 = 8*7*6*5*4 = 6720
9*8*7 / 2! / 3!
There are three that I can see, there's clip, board and lip.
The only five letter word that can be made with those letters is 'ditto'.Other words that can be made with the letters in 'ditto' are:dodotIidittotot
There are ten letters BUT there are 3 Ss and Ts and 2 Is. So the answer is 10!/(3!*3!*2!) = 50400
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.
420.
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.
The number of 5 letter arrangements of the letters in the word DANNY is the same as the number of permutations of 5 things taken 5 at a time, which is 120. However, since the letter N is repeated once, the number of distinct permutations is one half of that, or 60.
5040
In a "language" containing just 10 letters, there are 10,000 four-letter permutations. It's easy to work out if you simply replace the letters with decimal digits 0-9. The first permutation is 0000, followed by 0001, 0002, 0003, ..., 9997, 9998 and 9999.