Finding how many odd numbers we can get from a given set of numbers We have a set of numbers: $\{0,2,3,5,6,8\}$
1.)We want to find the number of ways that we can make a 3 digit number.
2.)We want to know how many of those numbers from 1.) are odd.
3.)How many of numbers from 2.) can be divided by $3$ without a remainder.
4.) what if we want to create a number that is $n$ digits long, in how many ways can we do that (with repetitions)?
How I tried to solve it:
1.) $$\binom{5}{2}\binom{1}{1}\cdot2^2+\binom{5}{3}=50$$
So we took 2 numbers from 5 that are not zero, then we took 0 and we can put zero on  the second or third place, that is why I multiplied with 2. Then we can take the number that switched places with zero and put it in the first place. Then the number that was before in the first place switches places with zero, so I multiplied again by 2, and got $2^2$. I summed it with all possible combinations of numbers that are not zero. However the result is incorrect as I should do $6^3-6^2$ or $5\cdot6\cdot6$, but I still do not understand, where my understanding to do it with binomial symbols fails.
2.)Odd numbers from that set are: 3 and 5, if they are put last, that means that the whole number is odd. I think it is equally likely that any number from that set is last, thus: $(1.)\cdot\frac{2}{6}$. Where $(1.)$ is the result from the first problem.
3.) I don't know how to do this one... I know that we need to see if the sum is divisible by $3$, but I don't know how to do that with combinatorics.
4.)This is just $5 \cdot 6^{n-1}$
 A: For the first question, your work assumes that the digits cannot repeat whereas the official answer you are referring to assumes they can repeat. Even if the digits cannot repeat, the second term should be $ \displaystyle {5 \choose 3} \cdot 3!$. You did not multiply by $3!$ to permute the chosen digits.
If the digits can repeat, every digit in the three digit number has $6$ choices from the given set except the leftmost digit (hundredth place) cannot have zero. That leads to $5 \cdot 6 \cdot 6 = 180$ numbers.
For odd numbers with repetition allowed, there are $2$ choices for the one's place, $6$ choices for the tenth place and $5$ choices for the hundredth place. That leads to $2 \cdot 6 \cdot 5 = 60$ odd numbers.
For odd numbers divisible by $3$, the sum of digits should be divisible by $3$ too. If the numbers have $3$ in the one's place, the only possibility I see is $0, 3, 6$ in tenth place and $3, 6$ in hundredth place. For numbers with $5$ in one's place, both tenth place and hundredth place have choices of $2, 5, 8$.
A: *

*You have to choose 3 digits and you have a conditions. Number must be odd, that means it ends with 3 or 5. You have 2 options to choose the last digit, after it you have 5 options to choose the first digit(it can be 2,3,5,6,8 but not 0) and 6 options to choose the last digit in the middle.
$$5*6*2=60$$.

*Number can be divided by 3 if sum of it's digits is divisible by 3.
the lowest number is 111 and it's divisible by 3, then every next third number is odd and divisible by 3.
$$60:3=20$$
