Question 325397
<pre><b>
You can choose the first letter as any of the 7 letters in the word.  

For each of those 7 ways to choose the 1st letter, you can choose the 2nd
letter as any of the 6 remaining letters which weren't chosen for the first
letter.  That's 7*6 ways to choose the first 2 letters.

For each of those 7*6 ways to choose the 1st 2 letters, you can choose the 3rd
letter as any of the 5 remaining letters which weren't chosen for either of the
first 2 letters.  That's 7*6*5 ways to choose the first 3 letters.

For each of those 7*6*5 ways to choose the 1st 3 letters, you can choose the 4th
letter as any of the 4 remaining letters which weren't chosen for any the first
3 letters.  That's 7*6*5*4 ways to choose the first 4 letters.

For each of those 7*6*5*4 ways to choose the 1st 4 letters, you can choose the
5th, or last letter as any of the 3 remaining letters which weren't chosen for
any the first 4 letters.  That's 7*6*5*4*3 ways to choose the 5 letters.
 
Answer 7*6*5*4*3 = 2520 possible 5-letter passwords.

This can also be written 7P5 and the formula is

{{{N_POSITION_R}}}{{{""=""}}}{{{NPR}}}{{{""=""}}}{{{N!/(N-R)!}}}
{{{7_POSITION_5}}}{{{""=""}}}{{{7P5}}}{{{""=""}}}{{{7!/(7-5)!=7!/2!}}}{{{""=""}}}{{{(7*6*5*4*3*2*1)/(2*1)}}}{{{""=""}}}{{{(7*6*5*4*3*cross(2*1))/(cross(2*1))}}}{{{""=""}}}{{{7*6*5*4*3}}}{{{""=""}}}{{{2520}}}

OR if you prefer

{{{N_POSITION_R}}}{{{""=""}}}{{{N*(N-1)*(N-2)*"..."}}} --> (R factors)

{{{7_POSITION_5}}}{{{""=""}}}{{{7*6*5*4*3}}} --> (5 factors) = 2520.


Edwin</pre>