Bunuel
1 red flag, 3 white flags and 2 blue flags are arranged in a line such that: no two adjacent flags are of the same colour and the flags at the ends are of 2 different colours. In how many different ways the flags be arranged?
A. 48
B. 10
C. 6
D. 4
E. 2
There are 6 flags and as adjacent flags have to be different, 1st and 6th are the conditions that need to be fulfilled first.
Since three color are there, following are the ways that condition can exist:
1. RW or WR
2. RB or BR (because it invalidates the condition - if BR/RB comes together there are combinations where 2 white flags would occur)
3. BW or WB
The inner flags can be arranged in many ways in the third case such that it has 4 ways.
Now, flags can be arranged(keeping all different colored flags together) in -
RWBWBW i.e. 1*3*2*2*1*1 = 12
OR
BWBWRW i.e. 2*3*1*2*1*1 = 12
OR
WBWRWB i.e. 2*3*1*2*1*1 = 12
But there are reverse of each of the cases that are also exact copies of the original sequence.
Hence uniquely arranged cases are \(\frac{12}{2} = 6\)
Answer C.