Use \(Total - Restriction\) technique.

If she chooses Green and Yellow together then there will be three choices for the third color, so she can choose {Green, Yellow, Any of the remaining three} in 3 ways,

The total number of ways to choose 3 out of 5 is 5C3 = 10.

\(Total - Restriction=10-3=7\)

Answer: A.
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The posters above me have demonstrated the approach that I'd typically use.
However, it's important to note that, when the answer choices are so small (as they are here), we should also consider the straightforward strategy of listing and counting

Let R, B, P, G and Y represent the colors Red, Blue, Purple, Green and Yellow respectively.

Now let's list the possible outcomes that meet all of the given conditions:
- RBP
- RBG
- RBY
- RPG
- RPY
- BPG
- BPY
Done!! So, there are 7 possible outcomes

Answer: A

Cheers,
Brent
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There are three cases: 1) green is one of the five colors chosen, but yellow isn’t, 2) yellow is one of the five colors chosen, but green isn’t, and 3) neither green nor yellow is chosen. Let’s analyze each case.

Case 1: Green is one of the five colors chosen, but yellow isn’t.

If green is chosen but yellow isn’t, then we have to choose 2 more colors from the 3 remaining colors. The number of ways to do that is 3C2 = 3.

Case 2: Yellow is one of the five colors chosen, but green isn’t.
This is analogous to case 1, so there are 3 ways for this case..

Case 3: Neither green nor yellow is chosen.

If neither color is chosen, then we have to choose 3 colors from the 3 remaining colors. The number of ways to do that is 3C3 = 1.

Thus, the total number of ways Flor can choose the colors for her project is 3 + 3 + 1 = 7.

Alternate Solution:

We can use the formula:

Total number of ways to pick 3 colors = number of ways where yellow and green are both included + number of ways where yellow and green are not both included

Since we are choosing 3 colors from 5 available colors, there are 5C3 = (5 x 4)/2 = 10 ways of doing this when there are no restrictions.

The number of ways where yellow and green are both included can be found easily by observing that yellow and green occupy two of the three slots; any one of the remaining three colors can occupy the final slot. So, there are 3 ways to choose colors where yellow and green are both included.

Thus, the number of ways to pick colors where yellow and green are not included together is 10 - 3 = 7.

Answer: A
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# select colors without green and yellow = #selecting 3 colors out of 5 - # selecting 1 color in 3(assuming green and yellow are picked) = 5C3 - 3C1

Sent from my Moto G (5) Plus using GMAT Club Forum mobile app
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RENAME colors to "unblock your brain": A, B, C, D and E.

Restriction: A and B cannot be BOTH chosen.

? = Number of choices of 3 colors among the 5 given, restriction obeyed.


First Scenario: neither A nor B is chosen.
There is just one possibility : CDE

Second Scenario: A is chosen (hence B is not)
There is just three possibilities: ACD, ACE, ADE.

Third Scenario: B is chosen (hence A is not)
This is similar to the previous scenario, hence additional 3 cases.

All cases mentioned above are MUTUALLY EXCLUSIVE, therefore they may be added: 7 possibilities.

All 7 cases are EXHAUSTIVE, hence we are sure the answer is (at least 7 and) not greater than 7.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Bunuel fskilnik GMATPrepNow

Please help

What am i doing wrong ?

1. Green But No Yellow = [1 (G) x 3 (1 of three other colors) x 2 (1 of two remaining colors ) ] / [3! (order does not matter)] + 2. Yellow But No Green = same as 1. + 3. Neither Green Nor Yellow = [3 (One of three colors other than G and Y) x 2 (One of two colors other than G and Y) x 1 ] / [3!] = 1 + 1 + 1 = 3

Number of ways to select 3 Colors out of 5 = 5C3 = 10

If Green and Yellow both are selected, number of ways to choose 1 color out of remaining 3 = 3C1 = 3

Number of ways to select 3 Colors out of 5 such that Green and Yellow both are not selected together = 10 -3 = 7

IMO A

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