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A computer password is required to be 9 characters long. How many passwords are possible if the password requires 3 letter(s) followed by 6 digits (numbers 0-9), where no repetition of any letter or digit is allowed?

I thought about using the formula: p(n, r) = n!/(n-r)!

So p(9,9) is wrong.. Not sure what to do for this problem..

4 Answers 4

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It should be $P(26, 3) \times P(10, 6)$

Or

You have 26 options to choose first letter.

Then remaining 25 letters. You have 25 options to pick.

For 3rd letter 24.

Similarly for first number 10 options, for second 9 ......., for 6th number 5 options.

We have $26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5$.

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    @Arthur actually I don't know how to write permutation in that form. I am trying that. But not work. So now I write it as this. – 2017-01-20
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the number of ways will be$P(26,3)×P(10,6)$

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$P(n, r)$ is the right tool to use; it's just not applied correctly. You need to first do the three letters, and then the six digits.

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You have

$$\#(\text{valid passwords}) = \#(\text{way to arrange 3 letters}) \cdot \#(\text{ways to arrange 6 digits})$$

each of the two quantities on the right can be computed with the $P(n, r)$ you were thinking about.