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Bane had 3 different color paints with him - Red, Green, and Blue. He wanted to paint a wall with 6 vertical stripes, b

Expert replies
by VJesus12 » Thu Mar 11, 2021 12:25 pm

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Difficulty

Bane had 3 different color paints with him - Red, Green, and Blue. He wanted to paint a wall with 6 vertical stripes, but no two adjacent stripes could be of the same color. Assuming that Bane can use one color more than once, in how many ways can Bane paint the wall?

A. 32
B. 64
C. 96
D. 243
E. 729

Answer: C

Source: e-GMAT
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Source: — Problem Solving |

VJesus12 wrote:
Thu Mar 11, 2021 12:25 pm
Bane had 3 different color paints with him - Red, Green, and Blue. He wanted to paint a wall with 6 vertical stripes, but no two adjacent stripes could be of the same color. Assuming that Bane can use one color more than once, in how many ways can Bane paint the wall?

A. 32
B. 64
C. 96
D. 243
E. 729

Answer: C

Source: e-GMAT
Take the task of painting the 6 stripes and break it into stages.

Stage 1: Select a color for the first stripe
Since we have 3 colors to choose from, we can complete stage 1 in 3 ways

Stage 2: Select a color for the 2nd stripe
This stripe cannot be the same color as stripe #1.
So, there are 2 remaining colors from which to choose, which means we can complete this stage in 2 ways.

Stage 3: Select a color for the 3rd stripe
This stripe cannot be the same color as stripe #2.
So, there are 2 remaining colors from which to choose, which means we can complete this stage in 2 ways.

Stage 4: Select a color for the 4th stripe
Applying the logic we applied above, we can complete this stage in 2 ways

Stage 5: Select a color for the 5th stripe
We can complete this stage in 2 ways

Stage 6: Select a color for the 6th stripe
We can complete this stage in 2 ways.

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus paint all 6 stripes) in (3)(2)(2)(2)(2)(2) ways (= 96 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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VJesus12 wrote:
Thu Mar 11, 2021 12:25 pm
Bane had 3 different color paints with him - Red, Green, and Blue. He wanted to paint a wall with 6 vertical stripes, but no two adjacent stripes could be of the same color. Assuming that Bane can use one color more than once, in how many ways can Bane paint the wall?

A. 32
B. 64
C. 96
D. 243
E. 729

Answer: C

Source: e-GMAT
Using the rules mentioned in the question STEM, we have

The first strip can be painted with 3 colors.
The second strip can be painted with 2 colors.
The third strip can be painted with 2 colors.
The fourth strip can be painted with 2 colors.
The fifth strip can be painted with 2 colors.
The sixth strip can be painted with 2 colors.

Hence number of ways \(\Rightarrow 3\cdot2\cdot2\cdot2\cdot2\cdot2 = 3\cdot2^5 \,\Rightarrow\, 3\cdot 32 = 96\) ways.

Therefore, C
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