[Math] In how many different ways can the flags be arranged in a line

Question

One red flag, three white flags and two blue flags are arranged
in a line such that:

$a.)$ No two adjacent flags are of the same color

$b.)$ The flags at the two ends of the line are of different colors
In how many different ways can the flags be arranged?

$\color{green}{a.)\ 6 }\\
b.)\ 4 \\
c.)\ 10 \\
d.)\ 2$

I did

Total ways$-$when same flags are considered as one
$-$flags at the two ends of the line are of different colours[w-w,b-b]

$=\left(\dfrac{6!}{3!\times 2!}\right)-(3!)-(2)=52$

which is not in options.

I look for a short and simple way.

I have studied maths upto $12$th grade.

Answer 1

Hint:

If you do not start or end the line with a white flag then it is unavoidable that $2$ white flags will be adjacent.