How many differrent ways can you arrange the letters of the word revolver?

Answer 1

Revolver has eight letters. Assuming you do not care if you make words, there are 40,320 permutations for arranging those eight letters, if you do not count the repeated e's and v's and r's

8! = 40,320 = (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8)

We must take this result and divide out the repeated elements. Since there are 2 e's, 2 v's and 2 r's:

40,320/(2!2!2!) = 5040.

Answer 2

25,200 different letter arrangements.

You have 8 letters where 3 are repeated. Let's represent them as 2 groups of letters:

A, B, C, and A, B, C, D, E.

The number of different 3 letter permutations that the first group of letters can make, A, B, C, is: 3! = 6

The number of different 5 letter permutations that the second group of letters can

make, A, B, C, D, E, is: 5! = 120.

From the 8 total spaces that the letters ocupy, the 3 letters of the first group can only

ocupy 7, because they can only take the space of one side of the letter equal to they.

Ocupying the other side is equivalent to a permutation between them and that does

not produce a new array.

So the number of different arragements of letters that can be made in the word

revolver is: 3!∙ 5!∙ 7C3 = (3x2)∙(5x4x3x2)∙[7!/(3!∙4!)] = 25,200

If all the 8 letters where different, the number of 8 letter different arrangements possible is 8! = 40,320