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

  1. no two adjacent flags are of the same colour.
  2. the flags at the two ends of the line are of different colours.

In how many different ways can the flags be arranged?

  1. $6$
  2. $4$
  3. $10$
  4. $2$
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Red flags are represented by R

White flags are represented by W

Blue flags are represented by B

And vacant spaces are represented by _

As No 2 adjacent flags are of the same colour, these are the 2 possible arrangements-

       1.  W_W_W_

OR

       2. _W_W_W

Now, 1 red flag and 2 blue flags have to be arranged in these vacant places.

 Hence, these 3 flags can be arranged in 3!/2! ways = 3 ways { RBB, BRB, BBR}

If we consider the 1st possible arrangement, then also these 3 flags can be arranged in 3 ways and this is applicable to the 2nd possible arrangement also.

So, Total number of ways = 3 + 3 ways

                                          = 6 ways. (option 1))

 

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