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# John has to paint a wall with seven horizontal stripes. He

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SVP
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John has to paint a wall with seven horizontal stripes. He [#permalink]  26 Oct 2008, 02:05
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

thanks
Manager
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Re: Permutation & Combination [#permalink]  26 Oct 2008, 04:16
tarek99 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

I'm getting (B) 210.
(i) Given the limitations, the only possible combination is 4 stripes of one colour, and 3 of another, therefore 6 colour schemes (4r3b, 3r4b, 4y3b, 3y4b, 4r3y, 3y4r)

(ii) To calculate the arrangements of stripes on the wall = 7c4 arrangements (you "arrange" the 4 stripes of one colour, and the 3 other stripes go into the other spaces)

(i)*(ii) = 6*35 = 210
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Re: Permutation & Combination [#permalink]  27 Oct 2008, 05:28
The OA is indeed B= 210. Would you please explain how you got the 35? Let me share with you another way that I found and please let me know whether this approach is considered correct:

First of all, we have only 3 different types of colors available. These colors are red, blue, and yellow. So out of these 3 colors, we will need to see how many ways we can select 2 colors, therefore: 3!/(1!*2!) = 3 different ways of selecting 2 colors from our total option of only 3 colors.

Next, out of these 3 ways _ _ _

how many arrangements of different horizontal strips can we place in each way.

In the first way, we have 7, in the second, we have 6, and in the third, we have 6, therefore:

7 * 6 * 5 = 210

is this correct approach?
Manager
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Re: Permutation & Combination [#permalink]  27 Oct 2008, 06:10
@tarek99:
I got 35 by doing 7C4: Out of the 7 total possible stripes, I choose the position of the 4 coloured stripes, and the remaining 3 stripes are of the other colour.

------

I agree with the first part, there are 3 possible ways to choose 2 colours: Red+Blue, Blue+Yellow, Yellow+Red [in the format A+B]

I didn't understand the second part - not sure why you would multiply those values.
If I followed your first part, my second part would be as follows:

For each of the 3 unique colour combinations above, there are 2 unique ways to use them: either [4 of colour A + 3 of colour B] or [3 of colour A + 4 of colour B]. This is given in the question ("maximum 4 stripes of 1 colour"). Therefore multiply by two.

The last part would be, for each of the combinations above, how many unique ways are there to arrange the stripes on the wall? This part is the same as my solution, 7C4 = 7!/(4!*(7-4)!) = 7! / 4!3! = 7*6*5/3*2 = 7*5 = 35 unique ways to arrange the stripes.

The total solution is thus then 3 * 2 * 35 = 210.

Does that make sense? Sorry I didn't understand your way though, it doesn't mean it isn't just as good
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Re: Permutation & Combination [#permalink]  28 Oct 2008, 23:25
Explanation from prince is more clear. I do not know where 7*6*5 is coming from in the OE.
Re: Permutation & Combination   [#permalink] 28 Oct 2008, 23:25
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