A ship has a flagpole which has 3 different positions in which it can fly a flag. Assume the ship carries four flags of which two are identical solid red and two are identical solid blue.
(a) How many different one flag signals could be displayed? (Assume that the same flag, flown in a different position, is a different signal.)
(b) How many different two-flag signals could be displayed?
(c) How many different signals (including no flags at all) could be displayed?
My attempt so far:
Firstly it seems these all deal with permutations since order matter.
$$ \frac{n!}{(n-r)!} $$
a) I have this to to be equivalent to $ \frac{3!}{(3-2)!}$ as 3 is the number of positions and there are two potential flag colors. Fairly straight forward as 6 permutations.
b) This one I'm confused how to construct. Intuitively $ 3!2! $ makes sense as the number of positions the two flags can occupy multiplied by the two color combinations. It also adds up to 12, which is the answer if I sketch it out.
c) No idea with this one. Too many overlapping factors to consider. Doubting that it can be expressed in a single permutation.
Best Answer
For $(a)$, there are $3$ positions to choose and for each position, there are $2$ colors available. Thus the answer is $3 \cdot 2 = 6$.
For $(b)$, there are $\binom{3}{2}$ position combinations and for each position combination, you can choose $4$ color combinations (i.e., RR, RB, BR, BB), thus there are totally $3 \cdot 4 = 12$ ways.
For $(c)$, for convenience, we temporarily suppose that there are $3$ red flags and $3$ blue flags. We count the total # of ways under this assumption and then remove those cases that use $3$ flags of same color. For each position, we have three options: red, blue and empty. Therefore, the total # of ways is $3^3 = 27$, in which there are two cases where we use all red flags or all blue flags. Thus, we conclude that the answer is $27 - 2 = 25$.