DarkHorse2019 wrote:
GMATinsight wrote:
Kritesh wrote:
A secretary differentiates files by a variety of color combinations. If she uses 4 different colors, and marks each file with 1, 2, or 3 of the colors, how many different color combinations can she generate?
(A) 4
(B) 12
(C) 14
(D) 15
(E) 24
If she marks by only one color then total possible combinations = 4C1 = 4
If she marks by only two colors then total possible combinations = 4C2 = 6
If she marks by only three colors then total possible combinations = 4C3 = 4
Total possible combinations =
4+6+4 = 14
Answer: Option C
We usually multiply the combinations when combining the result, why did we add here?
Please help
chetan2u,
Gladiator59,
VeritasKarishma,
Bunuel,
generisPartial solutions are multiplied (A AND B AND C) and complete solutions are added (X OR Y).
Let's understand this with an example:
Say there is a fast food joint which serves meals consisting of a burger/pizza slice, a side and a drink.
There are 3 different kids of burgers, 4 different kinds of pizza, 2 different sides and 5 different drinks available.
In how many different ways can you make a meal?
Let's make burger meals first. We can choose a burger in 3 ways, sides in 2 ways and drinks in 5 ways.
You need to choose a burger AND a side AND a drink.
So total burger meals = 3*2*5 = 30 burger meals
Now let's make pizza meals. We can choose a pizza slice in 4 ways, sides in 2 ways and drinks in 5 ways.
You need to choose a pizza AND a side AND a drink.
So total pizza meals = 4*2*5 = 40 pizza meals
Now, to choose a meal, you pick either a burger meal OR a pizza meal. The 30 burger meals are complete in themselves and 40 pizza meals are complete in themselves too.
So you ADD them to get 30 + 40 = 70 meals. You can pick a meal in 70 ways.
In the original question, she can mark a file using 1 colour only. So the 4 colours give us 4 different complete solutions.
OR she can mark a file using 2 colours 4C2 = 6 ways of marking a file using 2 colours (complete solutions).
Or she can mark a file using 3 colours in 4C3 = 4 ways. Again complete solutions
So we add them to get 4 + 6 + 4 = 14 solutions.
_________________
Karishma
Veritas Prep GMAT Instructor
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