1 red flag, 3 white flags and 2 blue flags are arranged in a line such

Question

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

A. 48
B. 10
C. 6
D. 4
E. 2

gmatbyexample · Accepted Answer

The other approach is to think in terms of White:
If there are 3 whites and they are never together - then out of 6, they will either take all odd places or all even places
White in odd places:
The other 3 remaining place will be taken by 1 Red and 2 Blue. So an arrangement of 3 with 2 things similar: 3!/2!

White in even places:
Exactly same logic: 3!/2!

Total: 3!/2! + 3!/2! = 6

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imeanup · Answer

Given information;
 R =1 
 W =3 
 B =2

Case 1:
 R - - - - \frac{W}{B}  Only  W  is possible in the last position. If  B  takes the last position then it violates the condition.

Case 2:
 W - - - - \frac{R}{B}  possible combinations are  W B W B W R  ;  W R W B W B  ;  W B W R W B

Case 3: 
 B - - - - \frac{R}{W}  possible combinations are  B W B W R W   B W R W B W

IMO  Ans C

unraveled · Answer

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.

dictanobis · Answer

Rule 1: No two balls of the same colour can be adjacent.
Rule 2: The starting and ending can not be the same colour.

Let's work by handling the "Elephant in the room" first - White (Since they are the highest in number)

W _ W _ W - With this, we have accounted for 5 places out of 6.

The 6th can come in first place or last place (This is because if we put 2 blanks between any of the whites, rule 2 will be violated).
Hence, there are 2 ways we can arrange all 6 balls:

_ W _ W _ W

OR

W _ W _ W _

Now we need to fill these 3 empty spaces with 3 balls: 2 Blue and 1 Red.

This can be done in 3!/2! = 3 ways.

So number of possible arrangements: 2 * 3 = 6 ways
  Ans: C

1 red flag, 3 white flags and 2 blue flags are arranged in a line such

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1 red flag, 3 white flags and 2 blue flags are arranged in a line such

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