Q.How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 4?

659942 has the digit multiset $\{6,5,9,9,4,2\}$ — 9 is repeated twice, the rest distinct, and none of the digits is 0. Total distinct 6-digit numbers $=\dfrac{6!}{2!}=\dfrac{720}{2}=360$. For divisibility by 4, the two-digit number formed by the last two digits must itself be divisible by 4. Testing every distinct ordered pair from the digit set, the two-digit endings divisible by 4 are: 24, 52, 56, 64, 92, 96. For the endings 24, 52, 56, 64 (which use only one of the two 9's, leaving both 9's — wait, leaving one 9 and the other three distinct digits — in the front four positions): the front 4 digits are all distinct (only a single 9 remains among them alongside the other leftover digits), giving $4!=24$ arrangements each, so $4\times24=96$... but two of the 9's could also both remain: for endings 24, 52, 56, 64 the two digits used are non-9, so BOTH 9's remain among the front four digits, and the front is …

Compute the total via the identical-objects formula, then for the divisibility sub-condition test each possible last-two-digit ending against the divisibility-by-4 rule, and for each valid ending count the front-4-digit permutations — carefully checking whet …

Forgetting to check, for each ending separately, how many copies of the repeated digit 9 remain in the front four positions — t …