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probability

Expert replies
Source: — Problem Solving |

by Mike@Magoosh » Sun Jan 22, 2012 11:17 am
Hi, there. I'm happy to help with this. :)

I believe the complete text of the question is:
Six colours (red.black.white.orange.pink.yellow) can be used to decorate. If one or more can be used, how many ways are possible that white is used?
A. 30
B. 32
C. 26
D. 400
E. 720


Well, we have to consider six cases:
(i) one color used --- there's only 1 way for that color to be white.
(ii) two colors used --- there are five other colors that can be paired with white, so in other words, 5 pairs that contain white.
(iii) three colors used --- if white is used, it will be white plus a pair chosen from five, which is calculated 5C2 = (5!)/[(2!)(3!)] = (5*4)/2 = 10. There are 10 possible triplets with white.
(iv) four colors used --- if white is used, it will be white plus a trio chosen from five, which is calculated 5C3 = (5!)/[(3!)(2!)] = (5*4)/2 = 10. There are 10 possible quartets with white.
(v) five colors used --- that means, only one of the six omitted. There's one way to omit white, and 5 ways to omit a color other than white. Thus, there are 5 quintets that contain white.
(vi) six colors used -- only one possibility, which includes white. Thus, 1 set.

Take the sum: 1 + 5 + 10 + 10 + 5 + 1 = 32

Answer = B

Does that make sense? Please let me know if you have any questions on this.

Mike :)
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by mankey » Sun Jan 22, 2012 11:31 am
IMO: 2^5=32.

What is the OA?

Thanks.
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by Brent@GMATPrepNow » Sun Jan 22, 2012 11:32 am
sud21 wrote:Six colours (red.black.white.orange.pink.yellow) can be used to decorate. If one or more can be used, how many ways are possible that white is used?
Mike's approach is perfect.
Here's another approach . . .

It's often useful to list a few possible outcomes (in your head even) to see if any patterns develop.

Here's one possible outcome:
red - used
black - not used
white - used
orange - used
pink - not used
yellow - not used

Here's another possible outcome:
red - not used
black - not used
white - used
orange - not used
pink - not used
yellow - used

Now recognize that we can take the task of selecting various colors and break it into stages:
Stage 1: determine whether or not to use red
Stage 2: determine whether or not to use black
Stage 3: determine whether or not to use white
.
.
.
Stage 6: determine whether or not to use yellow


Stage 1 can be accomplished in 2 ways (use it or don't)
Stage 2 can be accomplished in 2 ways (use it or don't)
Stage 3 can be accomplished in 1 way (we must use it)
Stage 4 can be accomplished in 2 ways (use it or don't)
Stage 5 can be accomplished in 2 ways (use it or don't)
Stage 6 can be accomplished in 2 ways (use it or don't)

So, using the Fundamental Counting Principle, all six stages can be completed in 2x2x1x2x2x2 ways (=32 ways)

Cheers,
Brent
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by Anurag@Gurome » Sun Jan 22, 2012 5:25 pm
sud21 wrote:Six colours (red.black.white.orange.pink.yellow) can be used to decorate. If one or more can be used, how many ways are possible that white is used?
(1) White color can be used alone, which can be done in 1 way.
(2) White can be used with any other color, which can be done in 5C1 ways.
(3) White can be used with 2 other colors, which can be done in 5C2 ways.
(4) White can be used with 3 other colors, which can be done in 5C3 ways.
(5) White can be used with 4 other colors, which can be done in 5C4 ways.
(6) White can be used with 5 other colors, which can be done in 5C5 ways.

Therefore, number of possible ways that white is used = 1 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 1 + 5 + 10 + 5 + 1 = 32
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by GMATGuruNY » Sun Jan 22, 2012 8:10 pm
sud21 wrote:Six colours (red.black.white.orange.pink.yellow) can be used to decorate. If one or more can be used, how many ways are possible that white is used?
Given n elements, the number of ways to select 0 or more of the n elements = 2^n.

For example, given 3 people, the number of ways to select 0 or more of the 3 people = 2^3 = 8.
The reasoning behind the formula is that for each person there are 2 options: to be selected or not to be selected.
To combine the 2 options for each of the 3 people, we multiply:
2*2*2 = 2^3 = 8.

To confirm, here are all the ways to select 0 or more of the 3 people:
Number of ways to select 0 people = 1.
Number of ways to select 2 people = 3C2 = 3.
Number of ways to select 3 people = 3C3 = 1.
Total ways = 1+3+3+1 = 8.
Same answer as above.

In the problem at hand:
The white paint can be combined with 0 or more of the 5 other colors.
Thus, the number of combinations that include white = the number of ways to select 0 or more of the 5 other colors = 2^5 = 32.
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