There are two methods for finding this:

Method 1:



and 

Method 2:



Explanation of method 1:

Since "seventeen" is a 9-letter word, there are 9 positions 
to place the letters.

We can choose any 1 of the 9 positions for the "s". 
That's 9C1 ways to place the s. After placing it, that
leaves 8 positions to fill.

We can then choose any 4 of the 8 remaining positions 
for the 4 "e"'s. That's 8C4 ways to place the 4 e's. After 
placing them, that leaves 4 positions to fill.

We can then choose any 1 of the 4 remaining positions 
for the "v". That's 4C1 ways to place the "v". After 
placing it, that leaves 3 positions to fill.

We can then choose any 2 of the 3 remaining positions 
for the 2 "n"'s. That's 3C2 ways to place the 2 "n"'s. After 
placing them, that leaves only 1 position to fill.

So we can chose the position for the "t" in only 1C1 or 1 way.

So the answer using method 1, is



Explanation of Method 2:

First pretend that the 4 e's are all different colors, so that we 
could tell them apart. And also pretend that the 2 n's are different 
colors, so that we can tell them apart too.

That is let's pretend that the word "seventeen" is spelled with
colored letters, like this

seventeen

So there would be 9! or 362880 ways to arrange them if we could 
tell the e's apart and also tell the n's apart.

So one random sample of such an arrangment of "seventeen" is, 
say, this:

nevtesene

Now within this one particular sample arrangement there are 4! ways
we can arrange the colors of the e's and there are 2! ways we can
arrange the colors of the n's. 

So that's 4!2! or 48 ways to move the e's and n's around in that 
one particular sample. And every one of these would be a
case of the arrangement "nevtesene". 

Every sample would be the same, so that means that the 9! or 362880
counts every arrangement 4!2! or 48 times too many, so we must 
divide it by 48 to get the correct answer. So the answer, using 
method 2 is

 

Edwin