Answer:
Step-by-step explanation:
i feel like it would be (4)(4)(4)(4)(4)(4)(4)(4) = 65536
The number of different four digit passwords that are possible if we use letters and digits only, and can't include repeating characters is 13,388,280 ways.
We can use combinations for this case,
Total number of distinguishable things is m.
Out of those m things, k things are to be chosen such that their order doesn't matter.
This can be done in total of [tex]^mC_k = \dfrac{m!}{k! \times (m-k)!}[/tex] ways.
If the order matters, then each of those choice of k distinct items would be permuted k! times.
So, total number of choices in that case would be:
[tex]^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}[/tex]
This is called permutation of k items chosen out of m items (all distinct).
For this case, we assume that:
"Letters" are denoting uppercase and lowercase english alphabets (26 uppercase, and 26 lowercase, so total 52 such letters).
And digits are denoting (0,1,2,3,4,5,6,7,8,9) so total 10 such digits.
Thus, total 52+10 = 62 different choices to choose from.
Of these 62 options, we can use one letter only one time in a password.
Choosing any 4 characters out of those 62 characters is done in:
[tex]^{62}C_4 = \dfrac{62 \times 61 \times 60 \times 59 \times 58!}{4 \times 3 \times 2 \times 1 \times 58!} = 557845[/tex] ways.
Each of those 4 selections have distinct characters, and in a password, their arrangement matters, so each of those 557845 choices can be permuted in 4! ways.
Thus, total number of distinct 4 digit password out of 52 english letters and 10 digits is [tex]4! \times 557845 = 24 \times 557845= 13388280[/tex] ways. (or we could've directly evaluated [tex]^{62}P_4[/tex]
Thus, the number of different four digit passwords that are possible if we use letters and digits only, and can't include repeating characters is 13,388,280 ways.
Learn more about combinations and permutations here:
https://brainly.com/question/13387529
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