Combinations of $5$ letters and $3$ numbers in an $8$ digit password
I'm having a pretty hard time solving this relatively simple problem. Essentially, the problem asks how many 8 digit passwords can be formed using 5 different letters from a set of uppercase and lowercase letters (thus, $52$ letters in total) and $3$ different digits from the set of integers $0-9$. My solution so far is that there are $\binom{8}{5}\binom{52}{5}\binom{10}{3}$ possible passwords, but I have no way of checking is this is true and the other problems haven't come quite so easily to me. Am I correct? If not, how can I find the correct answer? Thank you!
$\endgroup$ 31 Answer
$\begingroup$It is said that $5$ distinct letters (uppercase and lowercase ) and $3$ distinct digits will be used to construct lenght of $8$. So , lets assume that there are $8$ spaces to fill with characters.
Lets choose $5$ spaces among $8$ spaces to place the letters by $\binom{8}{5}$ , so the rest will be for digits.
It is also said that letters and digits must be different , so we can use permutatiton. Permutation for letters is $P(52,5)$ , permutation for digits is $P(10,3)$
Hence , the result is $$\binom{8}{5} \times P(52,5) \times P(10,3) $$
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