Probability with colored flags

Question

A signal has $6$ flags, each flag can be blue, white or red. Possible signals formed is $n^r = 3^6 = 729$ possible signals formed

How many different signal can be made from $6$ flags of which $3$ are white, $2$ are red and $1$ is blue?

Order does matter - so permutation is this a case of $3! \cdot 2! \cdot 1!$ so $6 \cdot 2 \cdot 1 = 12$ ways? I'm not sure if this is right ...

Answer 1

The number of choices for the blue flag is $6$. For each of these choices, there are $$\binom{5}{2}$$ ways to choose the two red flags of the remaining five flags. The last three flags have to be white.

---

This gives a final answer of $$6 \cdot \binom{5}{2} = 6 \cdot \frac{5\cdot 4}{2} = \boxed{60}$$

Answer 2

Out of the six flags, there are $\binom{6}{3} = 20$ ways to choose which flags are going to be white. Now out of the remaining three flags, pick which one is going to be blue, and the remaining two will automatically be red.

Answer 3

First place the $3$ whites so that is $6 \choose 3$. Out of the $3$ remaining spots, place the $2$ reds so that is $3 \choose 2$. There is only $1$ remaining spot for the blue.

So the answer is $6 \choose 3$ * $3 \choose 2$ * $1$ = $60$.