1
$\begingroup$

Suppose a system requires you to choose a 5 character password. Only the digits 0-9, lowercase letters, and 3 special characters are allowed to be used, and no character is allowed to be repeated. If at least 1 special character is also required, how many possible 5 character passwords can there be?

So I know for this I have to use combinations and permutations but I'm not sure how to approach multiple options like this. I'll try to write out my thought process: (SC = special character)

a (1 SC) = nCr(3, 1) * nPr(36, 4) * nPr(5, 1)

(choose 1 SC out of 3) * (choose 4 out of 36 digits or letters, no repetition) * (choose 1 placement in the 5 characters for the SC)

b (2 SCs) = nCr(3, 2) * nPr(36, 3) * nPr(5, 2)

(choose 2 SC out of 3) * (choose 3 out of 36 digits or letters, no repetition) * (choose 2 placements in the 5 characters for the SCs)

c (3 SCs) = nCr(3, 3) * nPr(36, 2) * nPr(5, 3)

(choose 3 SC out of 3) * (choose 2 out of 36 digits or letters, no repetition) * (choose 3 placements in the 5 characters for the SCs)

a + b + c = answer

Am I on the right track here? Any feedback is appreciated!

$\endgroup$
3
  • 1
    $\begingroup$ One way is to find the number of passwords in total with no repeats and subtract the number of passwords with no repeats and no special characters $\endgroup$
    – Henry
    Oct 7, 2021 at 17:42
  • $\begingroup$ You have $$3\cdot 5$$ possibilities to choose the special character and its position. Letting $$n=\text{Number of allowed characters} = 10+26+3,$$ then you have $(n-1)(n-2)(n-3)(n-4)$ possibilities left. $\endgroup$
    – Maximilian Janisch
    Oct 7, 2021 at 17:42
  • 1
    $\begingroup$ @Henry ohhh so nPr(39, 5) - nPr(36,5), I can't believe I didn't think of that! $\endgroup$
    – Qing Zheng
    Oct 7, 2021 at 17:50

2 Answers 2

Reset to default
1
$\begingroup$

$N$ - number of all characters (specials included)
$n$ - number of non-special characters
number of all possible 5 letter words from all possible characters ${N\choose 5} 5!$
number of all possible 5 letter words from non-special characters only ${n\choose5} 5!$ $$\text{Answer: }{N\choose5} 5! - {n \choose 5} 5!$$

$\endgroup$
2
  • 1
    $\begingroup$ "no character is allowed to be repeated" $\endgroup$
    – Henry
    Oct 7, 2021 at 17:54
  • $\begingroup$ Thanks for the comment, changed it. $\endgroup$
    – Gábor Pálovics
    Oct 7, 2021 at 18:00
0
$\begingroup$

$\mathbf{\text{NOTE=}}$ I give this answer for showing what you can do for more advanced questions .

I see that you have a problem with how many number of special character , digit and letters must be and the order of them in password. Hence , in your approach , you are missing very important thing , it is the location of different types of integrents.

What i meant by "the location of different types of integrents." is that the password $L-D-SC-SC-D$ is different from $D-L-L-SC-L$ where $L,D,SC$ means letter, digit , special character , respectively.

Then , how we handle this situation ? We can calculate the situations exhaustively such that $1$ SC , $1$ D , and $3$ L etc. As you see ,this is very cumbersome even for $n=5$. Then , we need a shortcut to get rid of this torturus process .

So , who is our hero to save us from this swamp ? The answer is exponential generating functions.

Lets say that the exponential generating function of special characters is $$\bigg(x + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)$$ we start from $x$ and end in $x^3$ , because the repetition is not allowed and it must appear.

Now , we should add some coefficients to this exponential generating function , the coefficients will be the permutatitons of the characters ,according to the exponential such that $$\bigg(P(3,1)x + P(3,2)\frac{x^2}{2!} + P(3,3)\frac{x^3}{3!} \bigg)=\bigg(3x + 3{x^2} + {x^3} \bigg)$$

Lets say that the exponential generating function of letters is $$\bigg(1+x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg)$$ We start form $x^0 =1$ and end up $x^4$ , because there is not any restriction over it. Now , place the coefficients to find the permutatitons according to the exponential such that $$\bigg(P(26,0)1+ P(26,1)x + P(26,2)\frac{x^2}{2!} + P(26,3)\frac{x^3}{3!} + P(26,4)\frac{x^4}{4!}\bigg)$$

It is equal to $$\bigg(1+ 26x + 325{x^2} + 2600{x^3} + 14950{x^4}\bigg)$$

Lets say that the exponential generating function of digits is $$\bigg(1+x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg)$$ We start form $x^0 =1$ and end up $x^4$ , because there is not any restriction over it. Now , place the coefficients to find the permutatitons according to the exponential such that $$\bigg(P(10,0)1+ P(10,1)x + P(10,2)\frac{x^2}{2!} + P(10,3)\frac{x^3}{3!} + P(10,4)\frac{x^4}{4!}\bigg)$$

It is equal to $$\bigg(1+ 10x + 45{x^2} + 120{x^3} + 210{x^4}\bigg)$$

Now , find the coefficient of $x^5$ in the expansion of $$\bigg(3x + 3{x^2} + {x^3} \bigg) \times \bigg(1+ 26x + 325{x^2} + 2600{x^3} + 14950{x^4}\bigg) \times \bigg(1+ 10x + 45{x^2} + 120{x^3} + 210{x^4}\bigg) $$

After that , multiply it by $5!$ , it is the result.

So , $$5! \times 198765 =23,851,800$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .