Calculation of finding way to make a password

Question

A computer password requires you to use exactly 1 uppercase, exactly 3 lowercases, 3 digits and 2 special charecters(given that there are 33 special charecters that can be used)(every thing can be repeated) . In how many ways you can create password?

My attempt: 1 uppercase can be choosen by ${26 \choose 1}$ ways, 3 lowercases can be choosen by ${26\choose 1}$$\times {26\choose 1}$$\times {26\choose 1}$ ways, 3 digits can be choosen by ${10\choose 1}$$\times {10\choose 1}$$\times {10\choose 1}$ways and 2 special charecters can be choosen by ${33\choose 1}$$\times {33\choose 1}$ ways. But I am stucking for further progress. Please help me. Thanks in advance

The question is given exactly which I type. So I think it can be repeated — Debprasad Kundu, Mar 07 '21 at 06:49
does this https://math.stackexchange.com/q/739874 help you. I don't think so but you can try it — , Mar 07 '21 at 06:51
You need $3$ digits and $2$ special characters. Right? Probably you have messed these two things. — Anacardium, Mar 07 '21 at 07:08
@Anacardium we can choose by these way. But chosable things are again permuteted. Right? — Debprasad Kundu, Mar 07 '21 at 07:14
Ok. I edited that. But the question was exactly same in the paper from which I took. — Debprasad Kundu, Mar 07 '21 at 08:09
I understand. I was talking about where you posted the question from, that they should have been more explicit. Based on repetitions being allowed, your working is fine except you needed to also find positions for them. — Math Lover, Mar 07 '21 at 08:33

true blue anil · Answer 1 · 2021-03-07T07:39:47.573

2

You want a configuration like $\,U\,L\,L\,L\,D\,D\,D\,S\,S\,$ and you already know how many ways there are to choose for this configuration.

Now configurations can be permuted in $\dfrac{9!}{1!3!3!2!}$ ways, just multiply

edited Mar 07 '21 at 07:39

answered Mar 07 '21 at 07:15

true blue anil

41,295

score 1 · Answer 2 · answered Mar 07 '21 at 08:25

As the repetition is allowed, first find number of ways to choose positions of $1U, 3L, 3D$ and $2S$ ($U$ - uppercase, $L$ - lowercase, $D$ - digit, $S$ - special character).

That can be done in $\displaystyle 9 \cdot {8 \choose 3} \cdot {5 \choose 3}$ ways.

Now uppercase can be chosen in $26$ ways, lowercase can be chosen in $26^3$ ways, digits can be chosen in $10^3$ ways and special characters can be chosen in $33^2$ ways.

So total number of passwords of length $9$ that can be made

$ = \displaystyle 9 \cdot {8 \choose 3} \cdot {5 \choose 3} \cdot 26^4 \cdot 10^3 \cdot 33^2$

Calculation of finding way to make a password

2 Answers2