Question 1115873: You are making a new password. It must consist of 4 letters and 2 numbers. How many passwords can be created if no numbers or letters are repeated?
My work :
26*25*24*23*10*9 = 32292000
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! Your answer assumes case does not matter ("a"="A") and perhaps more importantly, it assumes a fixed arrangement of letters and numbers for the 6-character password.
If the password can have any arrangement of 4 letters and 2 digits, then you must multiply your answer by 6! to count the number of those arrangements (ABCD12, A1BDC2, 12DBAC, etc. <— 6! arrangements can be made from just one selection of 4 distinct letters and 2 distinct digits).
I hope this helps and that I haven't missed anything.
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EDIT: Tutor ikleyn may have missed this: "It must consist of 4 letters and 2 numbers."
I am not confident in my answer either. It seems by multiplying by 6! I am double counting some permutations. I am beginning to think the answer should be 26C4*10C2*6! = 484380000. In other words, choose the 4 letters from 26, then choose 2 digits, then shuffle the 6 characters to make a password. Note that the student has stated in a message to me that the letters and numbers can be in any order, so it is not necessarily four letters followed by two digits.
EDIT 4/28: I would say 26C4*10C2*6! is correct. Unlike many of the other password problems here on algebra.com, we are not forming the password directly. For this problem we must choose the letters and digits first, and then as a second step shuffle them into a password. Most other password problems here are of the form "N letters FOLLOWED by M digits" which lends itself to calculating everything in one step.
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
Any of 26 letters and/or 10 digits can be in the 1-st position: 26+10 = 36 opportunities.
Any of remaining 26+10-1 = 35 symbols can be in the 2-nd position: 35 opportunities.
. . . . And so on . . .
In all, 36*35*34*33*32*30 = 36P6 different passwords are possible, satisfying given conditions.
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