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# How many different ways...?

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How many different ways...? [#permalink]  25 Mar 2011, 14:12
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Here's the question:
Ye Olde Arte has a large supply of colored poster board. The colors are blue, green, beige, white, yellow, and red. How many different ways can the store package 5 poster boards?
A. 126
B. 231
C. 252
D.378
E. 462

I know the answer, but I don't know how they get this answer.
Thank you ^^
[Reveal] Spoiler: OA
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Re: How many different ways...? [#permalink]  28 Mar 2011, 16:54
Expert's post
Good question!

Here is probably not the shortest way:

5 colors: $$C^6_5 = 6$$ - 5 colors out of 6.
4 colors: $$C^6_4 * C^4_1 = 15*4 = 60$$ - 4 colors out of 6 and one of 4 colors are used twice.
3 colors: $$C^6_3 * (C^3_1 + C^3_1) = 20*6 = 120$$ - 3 colors out of 6 and one color of 3 for 3+1+1 combination and one color of 3 for 2+2+1 combination.
2 colors: $$C^6_2 * (C^2_1 + C^2_1) = 15*4 = 60$$ - 2 colors out of 6 and one color of 2 for 3+2 combination and one color of 2 for 4+1 combination.
1 color: $$C^6_1 = 6$$ - 1 color out of 6.

6+60+120+60+6 = 252
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Re: How many different ways...? [#permalink]  28 Mar 2011, 18:17
Hi Walker

one color of 3 for 3+1+1 combination and one color of 3 for 2+2+1 combination.

Regards,
Subhash
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CEO
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Re: How many different ways...? [#permalink]  28 Mar 2011, 19:36
Expert's post
If we have only 3 colors for 5 boards, there are only 2 possible options 3 (first color) + 1 (second color) + 1 (third color) and 2(first color) + 2 (second color) + 1 (third color). There is 3 different ways for each option.

For example, option 1 if we have red, green, blue:

red + green + blue + blue + blue
red + green + green + green + blue
red + red + red + green + blue

option 2:

red + red + green + green + blue
red + red + green + blue + blue
red + green + green + blue + blue

So, $$C^3_1$$ actually means that we choose one color out of 3 for 3 boards (option 1) or for 1 board (option 2)
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Re: How many different ways...? [#permalink]  30 Mar 2011, 00:37
For 3 colors: one color of 3 for 3+1+1 combination and one color of 3 for 2+2+1 combination.

Can we write for the option 2 as 6C3 * 3C2 (which is equal to 6C3 * 3C1) ?

This is because let us say we have chosen 3 colors for 3 boards, and now we need to choose 2 colors out of 3 for the remaining 2 boards.

So option is 6C3 * 3C1 + 6C3 * 3C2.
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Re: How many different ways...? [#permalink]  30 Mar 2011, 03:14
Expert's post
Yup, $$C^3_2$$ is correct too.
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Re: How many different ways...? [#permalink]  18 May 2011, 21:56
good question indeed.
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Re: How many different ways...?   [#permalink] 18 May 2011, 21:56
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