Number of possible ways to raise flags on poles

Question

Suppose we have $n$ flagpoles and $r$ flags. Suppose we want to raise all flags, where a pole is allowed to have more than 1 flag:

1. How many possibilities are there if all flags are distinguishable (order matters), order on the pole matters?
2. How many possibilities are there if all flags are red?
3. How many different possibilities if the flags were colored red $p$ and white $q$ with ($p+q=r$)

For $1$) I thought since we can hang more than $1$ flag and the flags are distinguishable for every flag we have $n$ possible poles so $$\#\text{possibilites}=n\cdot n\cdot n\dots =n^r$$ answer is wrong since order on the pole matters

For $2)$ This one is a bit easier we had a lemma about it $$\#\text{possibilites}=\binom{n+r-1}{r}$$

$3$) Here I'm not sure, let $x$ be the number of the chosen flags, and $y$ the number of those that are red $$\# \text{possibilites}=\binom{n}{x}=\sum_{y=0}^x\binom{n-p}{x-y}\binom{p}{y}$$

Is this correct or am I choosing the wrong model?

Answer 1

Your answer to 1 is correct if you don't care about the order of the flags on each pole. If you do, you should multiply by the factorials of the numbers of flags on each pole. I don't see an easy answer for that. For example, if $r=3,n=2$ you would get $2^3=8$ ways to distribute the flags to flagpoles. Six of those have two flags on one pole and one on the other while two have all three flags on the same pole. If order matters there are $2!\cdot 6+3!\cdot 2=24$ possibilities.

Your answer to 2 is fine

For 3 I would do the same as 2. You have $r$ flags in a row. Choose $q$ of them to be white, which gives $n \choose q$ positions for white flags in the row. Now you have ${n+r-1 \choose r}$ ways to put dividers to put them on the poles, so there are ${n \choose q}{n+r-1 \choose r}$ configurations. This counts different orders on a pole as different, contradicting our answer to number 1

The approach for 3 will fix number 1. Put the flags in order on the ground in one of $r!$ ways, then put dividers between them (allowing multiple dividers in each space) to define which flags go on which pole, giving $r!{n+r-1 \choose r}$ configurations where order on the flagpole matters. For $r=3,n=2$ this gives $3!{4 \choose 3}=24$ possibilities.

Answer 2

(1) and (2) look great. I'm not sure what you mean by "the number of the chosen flags" in (3) -- doesn't the question specify you raise all $r$ flags? I would do (3) by noting that you can choose how to raise the red flags and the white flags independently, and for each color, you solve by (2).