How many 8-character passwords can be created from 26 lowercase letters and 10 digits, assuming that in every password there must be as many letters and numbers as?

Question

Josiah Allen · Accepted Answer

I'll give it a go: #26^4 * 10^4 * 8!#

Reasoning that "as many letters and numbers as" means we need 4 letters and 4 numbers.

So you'll make 4 picks from each category (letters and numbers).

You can duplicate them (i.e., pick each character more than once.)

...so, for your 4 picks from the letters, you'll have

#26 * 26 * 26 * 26 = 26^4# possible selections.

For each of these, you'll have 4 picks from the digits. Similar reasoning, this works out to

#10^4# combinations of digits.

This means you'll have #26^4 * 10^4# possible combinations of 4 letters and 4 digits, and, for each of these, you now need to work out how many unique ways to arrange them.

You'll have 8 possible choices for the 1st character, 7 for the second, 6 for the third, etc. This works out to #8!# different orderings for each of your #26^4 * 10^4# combinations of letters and numbers. Multiply everything out for the answer. It's a big number.

Best I can do, but I'm far from the ultimate authority for this kind of analysis, so, as always:

GOOD LUCK

Nora Amin · Answer

undefined To create an 8-character password with an equal number of lowercase letters and digits, we can split the password into four sections: four lowercase letters and four digits.

First, we calculate the number of ways to choose 4 lowercase letters from the 26 available lowercase letters, which is given by the combination formula $ C(n, k) = \frac{n!}{k!(n-k)!} $. In this case, $ n = 26 $ (total number of lowercase letters) and $ k = 4 $.

$$ C(26, 4) = \frac{26!}{4!(26-4)!} = \frac{26!}{4!22!} $$

Next, we calculate the number of ways to choose 4 digits from the 10 available digits, which is also given by the combination formula. In this case, $ n = 10 $ (total number of digits) and $ k = 4 $.

$$ C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} $$

Since there are no restrictions on the order in which the letters and digits are arranged within their respective groups, we multiply these two values to find the total number of possible arrangements.

$$ \text{Total arrangements} = C(26, 4) \times C(10, 4) $$

$$ \text{Total arrangements} = \frac{26!}{4!22!} \times \frac{10!}{4!6!} $$

$$ \text{Total arrangements} = \frac{26! \times 10!}{4!4!22!6!} $$

$$ \text{Total arrangements} = \frac{26! \times 10!}{4!4! \times 22! \times 6!} $$

$$ \text{Total arrangements} = \frac{26 \times 25 \times 24 \times 23 \times 10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1 \times 4 \times 3 \times 2 \times 1 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} $$

After simplifying, we find the total number of 8-character passwords with an equal number of lowercase letters and digits.