Problem Solving

A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is


(1)12x81
(2)16 x 192
(3)20x 125
(4)24x216
(5)None of the above

harsh.champ wrote:A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is


(1)12x81
(2)16 x 192
(3)20x 125
(4)24x216
(5)None of the above

The first stripe can be any colour. To avoid 2 of the same in a row, the second stripe can be 3 different colours. The same reasoning applies for the remaining stripes.

So, we have 4*3*3*3*3*3 possible flags, which is the same as 12*81... choose (1).

IMO E. Here is my approach..

Since two adjacent stripes cannot be of same color, minimum number of colors needed for a flag is 2 and max is 4.

No. of ways of choosing 2 out of 4 = 4C2 = 6.
with two colors, first stripe can be any of the two colors and rest all can be of only one color.
2 *1 * 1* 1*1 *1 = 2.
total # of ways of choosing 2 and creating 6 stripes = 6*2 =12

No. of ways of choosing 3 out of 4 = 4C3 = 4
3*2*2*2*2*2 = 3*32
Total # of ways of choosing 3 colors and creating 6 stripes = 4*3*32 = 12*32

No. of ways of choosing all 4 colors = 4C4 =1
4*3*3*3*3*3 = 12*81

Adding 12(1 + 32 + 81) = 12 * 114.

IMO E. Is this correct?

ace_gre wrote:IMO E. Here is my approach..

Since two adjacent stripes cannot be of same color, minimum number of colors needed for a flag is 2 and max is 4.

No. of ways of choosing 2 out of 4 = 4C2 = 6.
with two colors, first stripe can be any of the two colors and rest all can be of only one color.
2 *1 * 1* 1*1 *1 = 2.
total # of ways of choosing 2 and creating 6 stripes = 6*2 =12

No. of ways of choosing 3 out of 4 = 4C3 = 4
3*2*2*2*2*2 = 3*32
Total # of ways of choosing 3 colors and creating 6 stripes = 4*3*32 = 12*32

No. of ways of choosing all 4 colors = 4C4 =1
4*3*3*3*3*3 = 12*81

Adding 12(1 + 32 + 81) = 12 * 114.

IMO E. Is this correct?

_____________
Hey Stuart and Ian,
I read the above soln. posted by ace_gre.I can't pick out the mistake.
Can you tell where he went wrong??It is very important to clarify the approach and thinking pattern for P & C questions.
Seeking help!!!

Ian Stewart wrote:

harsh.champ wrote:

ace_gre wrote:IMO E. Here is my approach..

Since two adjacent stripes cannot be of same color, minimum number of colors needed for a flag is 2 and max is 4.

No. of ways of choosing 2 out of 4 = 4C2 = 6.
with two colors, first stripe can be any of the two colors and rest all can be of only one color.
2 *1 * 1* 1*1 *1 = 2.
total # of ways of choosing 2 and creating 6 stripes = 6*2 =12

No. of ways of choosing 3 out of 4 = 4C3 = 4
3*2*2*2*2*2 = 3*32
Total # of ways of choosing 3 colors and creating 6 stripes = 4*3*32 = 12*32

No. of ways of choosing all 4 colors = 4C4 =1
4*3*3*3*3*3 = 12*81

Adding 12(1 + 32 + 81) = 12 * 114.

IMO E. Is this correct?

_____________
Hey Stuart and Ian,
I read the above soln. posted by ace_gre.I can't pick out the mistake.
Can you tell where he went wrong??It is very important to clarify the approach and thinking pattern for P & C questions.
Seeking help!!!

ace_gre's solution is double-counting some of the possibilities. For example, the '4 colors' case above also counts all the flags that could be made using only two or three colours.

______________
Thanks,I got it!!
But the main concern is that on numerous of occasions I have also committed similar mistakes of double-counting some of the possibilities.Is there any formal technique that can help me omit the errors like the one above??
I specifically liked the method by which Stuart had solved it but that approach didn't click to me.

Thanks Stuart and Ian for providing the solution and pointing out my mistake.
I tend to double count sometimes making the problem more difficult than it actually is.

Like Harsh mentioned this is not the first time I am double counting. If there are ways to determine or realize that I have double counted (other than brute logic) I would be very happy and very much interested!!

Thanks!

ace_gre wrote:Thanks Stuart and Ian for providing the solution and pointing out my mistake.
I tend to double count sometimes making the problem more difficult than it actually is.

Like Harsh mentioned this is not the first time I am double counting. If there are ways to determine or realize that I have double counted (other than brute logic) I would be very happy and very much interested!!

Thanks!

My experience with counting questions is that people tend to overthink them and try to force a formula or two to apply, whereas for the vast majority of them simple logic is a much quicker means of attack.

When counting, here are the big things to ask:

1) does order matter? If not, then we use combinations; if yes, then we use permutations.

If order doesn't matter and you use permutations, you'll almost certainly double count unless you factor out the duplicate possibilities.

2) are all the items distinct? If not, then you need to factor out duplicates; again, this often leads to multiple counting of the same possibility.

IMO A.

Y __ __ __ __ __ __ = 1*3*3*3*3*3
G __ __ __ __ __ __ = 1*3*3*3*3*3
B __ __ __ __ __ __ = 1*3*3*3*3*3
R __ __ __ __ __ __ = 1*3*3*3*3*3

So finally we get = 4*3*3*3*3*3 = 12*81