Question 1094921: You need to have a password with 4 letters followed by 4 even digits between 0 and 9, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your password?
Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website!
There are 26 letters for the first letter choice.
25 for the 2nd
24 for the 3rd
23 for the 4th
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There are 5 choices for the first digit { 0, 2, 4, 6, or 8 }
4 for the 2nd
3 for the 3rd
2 for the 4th
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So there are 26*25*24*23*5*4*3*2 = 43056000 ways (=choices) to make a password this way.
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Alt solution: since order is important, we can use the permutation formula P(n,k)= n!/(n-k+1)!.
Viewing it this way, there are P(26,4) ways to pick (arrange) the letters and P(5,4) ways to pick (arrange) the digits: P(26,4)*P(5,4) = 358800*120 = 43056000.
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