SOLUTION: Please help How many 5 letter passwords can be formed from the letters PAYMENT if no repetition of letters is allowed? Thanks in advance.

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Question 325397: Please help
How many 5 letter passwords can be formed from the letters PAYMENT if no repetition of letters is allowed?
Thanks in advance.

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How many 5 letter passwords can be formed from the letters PAYMENT if no repetition of letters is allowed?
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There are 7 distinct letters in payment.
# of arrangements of 5 of the leters is 7P5 = 7!/(7-5)! = 7*6*5*4*3 = 2520
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Cheers,
Stan H.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

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%22%22=%22%22NPR%22%22=%22%22N%21%2F%28N-R%29%21
7_POSITION_5%22%22=%22%227P5%22%22=%22%227%21%2F%287-5%29%21=7%21%2F2%21%22%22=%22%22%287%2A6%2A5%2A4%2A3%2A2%2A1%29%2F%282%2A1%29%22%22=%22%22%287%2A6%2A5%2A4%2A3%2Across%282%2A1%29%29%2F%28cross%282%2A1%29%29%22%22=%22%227%2A6%2A5%2A4%2A3%22%22=%22%222520

OR if you prefer

N_POSITION_R%22%22=%22%22N%2A%28N-1%29%2A%28N-2%29%2A%22...%22 --> (R factors)

7_POSITION_5%22%22=%22%227%2A6%2A5%2A4%2A3 --> (5 factors) = 2520.


Edwin