a three-character password consists of 2 different digits between 0 and 9 inclusive. and 1 letter of the English alphabet. the letter must appear as first or second character, how many different passwords are possible?
I assumed that they are not case sensitive. So there is 26 possible letters for the first alphabet, then there is 10 digits for the second place and 9 digits for the third place. Redo the steps again and assume the letter is in second place. 10*26*9. I used permutation 26P1 * 10P2 so (26*10*9)^2=2340^2=5475600
But this answer is incorrect. Can somebody point out where i went wrong? :)
Thanks in advance :)
$\endgroup$2 Answers
$\begingroup$Your first two steps are correct. But you can not permute the available letters.
The calculations should be done as:
1) Put letter at first place : $26*10*9$ to fill
OR
2) Put letter at second place: $10*26*9$ ways to fill
The total ways will be sum of the two ways = $26*10*9*2 = 180*26 = 4680$
Assuming only 26 English and 10 numeric digits.
$\endgroup$ 2 $\begingroup$You multiplied the numbers of possibilities by each other rather than adding the numbers of mutually exclusive possibilities to each other.
Reality check: The number of three character passwords using $36$ characters without restriction is $36^3=46,656$ - and certainly less than $1,000,000$ on a very crude estimate - $5$ million is obviously wrong.
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