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by sana.noor » Sun Jul 28, 2013 10:47 am
John has to paint a wall with seven horizontal stripes. He only has enough paint for four red stripes, four blue stripes, and four yellow stripes. If he can use at most two colors, how many different ways can he paint the wall?

(A) 420
(B) 210
(C) 84
(D) 35
(E) 7

OA is B
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by [email protected] » Sun Jul 28, 2013 11:14 am
Hi sana.noor,

This situation is rather similar to other questions that have been posted here that ask, for example "how many codes can you make with the letters in PEPPER?" The catch is what you have to do when you have duplicate entries.

Here, we have 3 colors to choose from, BUT we're only allowed to use 2 of them. Furthermore, we have to paint 7 lines, BUT we can only have 4 lines of any color, so we would need 4 of one color and 3 of another. So, we're going to have duplicate entries; we just have to account for all of them.

With red, blue and yellow, we can have these combos:
red and blue
red and yellow
blue and yellow

Let's use red and blue as an example (and then we'll triple the result to cover all of the combos)

We'll have either
4 red and 3 blue
4 blue and 3 red

With 7 lines of paint, this math will require a permutation AND we have to factor out the duplicates:

7! / (4!)(3!) = 35

So we 4 red and 3 blue lines, there would be 35 options
With 4 blue and 3 red lines, there would also be 35 options

Red and Blue would give us a total of 70 options

Now we triple that to account for every combo of colors....

Final Answer:[spoiler]210 (B)[/spoiler]


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by GMATGuruNY » Sun Jul 28, 2013 11:52 am
sana.noor wrote:John has to paint a wall with seven horizontal stripes. He only has enough paint for four red stripes, four blue stripes, and four yellow stripes. If he can use at most two colors, how many different ways can he paint the wall?

(A) 420
(B) 210
(C) 84
(D) 35
(E) 7

OA is B
Each color may constitute at most 4 stripes.
Thus, of the 7 stripes:
4 stripes must be of ONE of the 2 colors selected.
The remaining 3 stripes must be of THE OTHER color selected.

Number of ways to select 4 stripes from 7 options = 7C4 = (7*6*5*4)/(4*3*2*1) = 35.
Number of color options for these 4 stripes = 3. (Any of the 3 colors.)
Number of color options for the remaining 3 stripes = 2. (Either of the 2 remaining colors.)
To combine these options, we multiply:
35*3*2 = 210.

The correct answer is B.
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