SOLUTION: Can anyone help me with this? not sure if it's Conditional Probability, Independent Event or using the Bayes' formula problem. Any help is greatly appreciated. ******************

Question 205505: Can anyone help me with this? not sure if it's Conditional Probability, Independent Event or using the Bayes' formula problem. Any help is greatly appreciated.
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A 8-character password consists of three letters (from A to Z) followed by five numbers (from 1 to 9).
a) How many different passwords can be formed?
b) How many different passwords have no repeated number or letters?
Thanks so much
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
there are 26 letters in the alphabet from a to z
there are 9 numbers from 1 to 9
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the total possible of combinations of passwords would be:
26*26*26*9*9*9*9*9
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all letters can repeat and all numbers can repeat.
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if you want no repeated numbers and letters than the total number of combinations can be:
26*25*24*9*8*7*6*5
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to see how this works take a much smaller example.
allow 3 letters (abc)
allow 2 number (12)
for a 2 character 1 number password the number of total pssibilities is:
3 * 3 * 2 = 18
the possible combinations are:
aa1 aa2 ab1 ab2 ac1 ac2
ba1 ba2 bb1 bb2 bc1 bc2
ca1 ca2 cb1 cb2 cc1 cc2
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if you did not want any repeats the total number of combinations would be:
3 * 2 * 2 = 12
the possible combinations are
ab1 ab2 ac1 ac2
ba1 ba2 bc1 bc2
ca1 ca2 cb1 cb2
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the possible repeats were in the characters because we only had one position for the numbers.
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in your bigger problem you had 3 positions for the charactrers and 5 positions for the numbers
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the concept was the same however.
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