p&c q21

[Saumyojit]

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?

okay i broke it down and approached like this which is not a proper method as if the no of each flag were high or then i cannot break down like that as it will be time consuming also i am not using p&C


arrangement start with R
R W B W B W

RWBWWB not possible

arrangement start with W
last flag R or B

W B W B W R
W R W B W B
W B W R W B
arrangement start with B

B W B W R W
B W R W B W
total 6 arrangements

 

[Dr.Peterson]

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?

okay i broke it down and approached like this which is not a proper method as if the no of each flag were high or then i cannot break down like that as it will be time consuming also i am not using p&C


arrangement start with R
R W B W B W

RWBWWB not possible

arrangement start with W
last flag R or B

W B W B W R
W R W B W B
W B W R W B
arrangement start with B

B W B W R W
B W R W B W
total 6 arrangements

Why is this not a "proper method"? The numbers are not large, so you don't need to do any more.

But if you had a similar problem with larger numbers, what you have done here might be a good model for seeing what restrictions there are on arrangements (e.g. you didn't need to consider RB... because there are fewer B than W, and an odd number of spaces to fill with them). The main change would be that rather than list all possible arrangements, you would count arrangements by type. Doing that in this case would be a waste of time, because listing is so easy.

Each different set of numbers would likely raise different issues, so there may not be a general, extensible method. Here, having only one R has a huge effect.

By the way, what you call "P&C" is just two tools out of a large set available in combinatorics. There is no reason to expect to use those in every problem. The most powerful and flexible tool is simply "thinking".

 

[sky2rain]

R doesn’t matter much in this question, the approach is to look at how many ways W and B can form according to the restriction of no adjacent color be the same. Firstly, WBWBW is obvious, because they are all separated. Secondly, you can have at most one pair of adjacent color to be the same because you can use that 1 red flag to separate. Stating with W: WWBWB, BWBWW, and with B…. B doesn’t seem to work because whenever B is adjacent to one another at least one W will connect with another to form two pairs. So based on my analysis, for the first case WBWBW you can slide the red in all 6 slots between every, in front of all and behind all letters. For second case you can only have two formations that is to use the red to separate the WW. So 6+2=8.