Question

At a certain company, any valid password consists of four digits followed by two lower case letters. Determine how many passwords are possible if the first digit cannot be a 0, the last letter must be a vowel (a, e, i, o, u) and repetition of any letter or any digit is not permitted?

Answer 1

The Password will be of Type: D₁D₂D₃D₄X₁X₂

Where D = Digits and X = Letter

Repetition is not allowed i.e. A digit which is used once cannot be used again, same is for letters.

For the First Digit, Normally we would have 10 options i.e. from 0-9

But it is given that First Digit cannot be 0.

Therefore, for the First Digit, we would have 9 Options i.e. from 1-9

For the Second Digit, we cannot use the digit which is used in the First position

But we can include 0, as it is not included in the first option.

Therefore, For the Second Digit, we would have 9 Options

For the Third Digit, we cannot use the Digits used in The First Two Places.

Therefore, For the Third Digit, we would have 8 Options

For the Fourth Digit, we cannot use the Digits used in The First Three Places.

Therefore, For the Fourth Digit, we would have 7 Options

We have 26 Alphabets. So, for the First Letter i.e. X₁, we would have 26 Options

But it is said that the Last Digit should be a vowel.

So, for X₁, we would have 25 Options

We have 5 Vowels. So for second letter i.e. X2, we would have 5 options

Now, to get the Total Possible Passwords, we would simply Multiply the Options for individual digits and letters i.e. :

= 9*9*8*7*25*5

= 567000 Passwords

NOTE: Since, the Options (i.e. possibilities) for individual digits and individual letters are dependent on other digits and letters, so we are Multiplying to get the Total Possible Passwords and not adding them.

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