We have no zeros, one $1$, two $2$'s, three $3$'s, four $4$'s, five $5$'s, six $6$'s, seven $7$'s, eight $8$'s, and nine $9$'s.
How many $12$-digit numbers can we make using at least three $6$'s so that all of these numbers are divisible by $9$?
What is the total number of numbers that have exactly three $4$'s and at least two $7$'s? (Here, other digits can also be used).
What is the total numbr of numbers that contain the $1$ that we have, do not contain any $4$, and do not have more than $6$ digits?
What is the total number of numbers that have at least one $2$, exactly two $3$'s, and at most three $4$'s? (Here, other digits can also be used).
To be honest, I have some knowledge about divisibility, factorials, permutations, combinations,.... but I do not know how to solve these questions that have constraints.
Any help would be appreciated. THANKS!