Flag with stripes

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

Flag with stripes

by harsh.champ » Thu Feb 04, 2010 1:27 pm

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

Quote

Source: Beat The GMAT — Problem Solving | Thu Feb 04, 2010 1:27 pm

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Thu Feb 04, 2010 2:13 pm

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).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Quote

ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pm; Thanked: 26 times; Followed by:1 members

by ace_gre » Thu Feb 04, 2010 2:18 pm

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?

Quote

GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Thu Feb 04, 2010 2:33 pm

For the leftmost stripe, you have 4 choices of colour. For the next stripe, you have 3 choices of colour (you can't repeat the colour of the leftmost stripe). For each remaining stripe, again you have 3 choices, since you can't repeat the colour of the stripe to the immediate left. Multiplying our choices, we have 4*3^5 flags in total, which, looking at the answer choices, is equal to 12*81 (I have no idea why they would express the answer in that way).

edit - looks like Stuart beat me to it!

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

Quote

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Thu Feb 04, 2010 2:46 pm

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!!!

Quote

GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Thu Feb 04, 2010 2:52 pm

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

Quote

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Thu Feb 04, 2010 2:56 pm

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.

Quote

ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pm; Thanked: 26 times; Followed by:1 members

by ace_gre » Thu Feb 04, 2010 7:59 pm

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!

Quote

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Thu Feb 04, 2010 8:43 pm

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.

Quote

neelimareddym: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Fri Jan 29, 2010 2:27 am; Thanked: 1 times

by neelimareddym » Thu Feb 04, 2010 9:36 pm

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