Combinatorics question

[bushra1175]

Hi everyone. I hope I'm posting in the right place for this question.

So the question reads:


So far I've gotten as far as visualising the problem but I'm struggling to come up with an actual answer

I know that there are two cases:

case 1: there are 2 digits in the password so there are this many possible values that can go in each slot


case 2: there is only one digit


So how can I numerically consolidate these two cases?

 

[JeffM]

You count by considering mutually exclusive cases.

One digit

[MATH]ssssd[/MATH],

Number of possibilities

[MATH]26^4*10[/MATH].

Two digits.

[MATH]sdssd \ \lor \ ssdsd \lor sssdd[/MATH].

Number of possibilities

[MATH]\dbinom{3}{1} * 26^3 * 10^2[/MATH].

And the answer is?

Is it enough?

 

[Romsek]

place your 1st pw char. There are [MATH]26[/MATH] possibilities.
place your 5th pw char. There are [MATH]10[/MATH] possibilities.

choose how many numeric chars there will be in the 3 remaining pw chars. There are 2 possibilities, 0 and 1.

If there are 0 numeric chars then there are [MATH]26^3[/MATH] possibilities for the remaining 3 pw chars

If there is 1 numeric char then choose what pw char it occupies. There are 3 choices
Then there are [MATH]10\cdot 26^2[/MATH] possibilities on how to fill the 3 chars.

Combining all this we have

[MATH]26\cdot 10\left(26^3 + 10\cdot 3 \cdot 26^2\right) = 9842560[/MATH] different passwords