I understand the solution, but my question is why doesn't calculating, P6 = 10*36^5 and the same for P7 and P8, work? 10 for the digit and 36 for the alphanumeric?
you're guaranteeing that there is a digit by using the 10, but, there's nothing to stop there from being a digit in the other positions (36 choices) that's why you have to get all the combs in which there are digits and subtract all the ones where there aren't which leads to P6 = 36^6 - 26^6
If I think if an alternate solution I'll revisit this post, but for now I don't see why you wouldn't be satisfied with the provided one, knowing why yours doesn't work
如果 6 个位置总共有 k 个数字，则有 6 over k 种可能性选择这 k 个位置。这些 k 个位置中的每一个都可以填充一个数字，因此您有 10^k 种可能的数字和 26^(6-k) 字母的可能性。
因此，对于 k 位数字，您有 10^k * 26^(6-k) 种可能性。因此，由于您可以以 6 over k 的方式分配 k 数字，因此您有
sum(k from 1 to 5: (6 over k) * 10^ k * 26^(6-k)) = 36^6-26^6
总共有 36^6-26^6 种可能性。
If you do it like 36^5 * 10, that means "take the first five positions and put some random letters/digits there and fill the sixth (and only the sixth) position with a digit" - but your digit can be in each place.
Consider the following: If you set 5 places with letters/digits, then putting a digit in the sixth place yields 10 possibilities (the sixth position can then hold every number from 0 to 9), and if you put the digit in front of the letters, this would yield another 10 possibilities (then, the first place could contain every digit from 0 to 9), so by multipliing by 10 you forget some possibilities.
If you want to calculate it "the raw way", you could do the following (I do it with six places, you can adapt it to 7 or 8). Since a password must contain at least one digit, it can hold 1, 2, 3, 4, 5 or 6 digits, resp 5, 4, 3, 2, 1 or 0 letters.
If you have k digits in 6 places in total, there are 6 over k possibilities to choose these k places. Each of these k places can be filled with a digit, so you have 10^k possibilities for the digits and 26^(6-k) possibilities for letters.
So, for k digits, you have 10^k * 26^(6-k) possibilities. Thus, since you can distribute the k digits in 6 over k ways, you have
sum(k from 1 to 5: (6 over k) * 10^k * 26^(6-k)) = 36^6-26^6
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您忘记了该数字可以位于任何位置。
You're forgetting that the number can be in any position.
您可以使用 10 来保证有一个数字，
但是，没有什么可以阻止其他位置上的数字（36 个选择）
这就是为什么你必须得到所有有数字的梳子并减去所有没有数字的梳子
这导致 P6 = 36^6 - 26^6
如果我认为如果有替代解决方案，我会重新访问这篇文章，但现在我不明白为什么你会对所提供的解决方案不满意，知道为什么你的解决方案不满足不工作
you're guaranteeing that there is a digit by using the 10,
but, there's nothing to stop there from being a digit in the other positions (36 choices)
that's why you have to get all the combs in which there are digits and subtract all the ones where there aren't
which leads to P6 = 36^6 - 26^6
If I think if an alternate solution I'll revisit this post, but for now I don't see why you wouldn't be satisfied with the provided one, knowing why yours doesn't work
如果你像
36^5 * 10那样做，这意味着“取前五个位置并在那里放置一些随机字母/数字，然后用数字填充第六个（并且只有第六个）位置” -但你的数字可以在每个地方。考虑以下情况：如果您用字母/数字设置 5 个位置，那么将数字放在第六位会产生 10 种可能性（第六个位置可以容纳从 0 到 9 的每个数字），并且如果您将数字放在字母，这将产生另外 10 种可能性（那么，第一个位置可能包含从 0 到 9 的每个数字），所以通过乘以 10，你会忘记一些可能性。
如果你想“原始方式”计算它，你可以执行以下操作（我用六个位置进行计算，你可以将其调整为 7 或 8）。由于密码必须至少包含一位数字，因此它可以包含 1、2、3、4、5 或 6 位数字，或 5、4、3、2、1 或 0 个字母。
如果 6 个位置总共有
k个数字，则有6 over k种可能性选择这k个位置。这些k个位置中的每一个都可以填充一个数字，因此您有10^k种可能的数字和26^(6-k)字母的可能性。因此，对于
k位数字，您有10^k * 26^(6-k)种可能性。因此，由于您可以以6 over k的方式分配k数字，因此您有sum(k from 1 to 5: (6 over k) * 10^ k * 26^(6-k)) = 36^6-26^6总共有 36^6-26^6 种可能性。
If you do it like
36^5 * 10, that means "take the first five positions and put some random letters/digits there and fill the sixth (and only the sixth) position with a digit" - but your digit can be in each place.Consider the following: If you set 5 places with letters/digits, then putting a digit in the sixth place yields 10 possibilities (the sixth position can then hold every number from 0 to 9), and if you put the digit in front of the letters, this would yield another 10 possibilities (then, the first place could contain every digit from 0 to 9), so by multipliing by 10 you forget some possibilities.
If you want to calculate it "the raw way", you could do the following (I do it with six places, you can adapt it to 7 or 8). Since a password must contain at least one digit, it can hold 1, 2, 3, 4, 5 or 6 digits, resp 5, 4, 3, 2, 1 or 0 letters.
If you have
kdigits in 6 places in total, there are6 over kpossibilities to choose thesekplaces. Each of thesekplaces can be filled with a digit, so you have10^kpossibilities for the digits and26^(6-k)possibilities for letters.So, for
kdigits, you have10^k * 26^(6-k)possibilities. Thus, since you can distribute thekdigits in6 over kways, you havesum(k from 1 to 5: (6 over k) * 10^k * 26^(6-k)) = 36^6-26^6possibilities in total.