Hi sana.noor,

This situation is rather similar to other questions that have been posted here that ask, for example "how many codes can you make with the letters in PEPPER?" The catch is what you have to do when you have duplicate entries.

Here, we have 3 colors to choose from, BUT we're only allowed to use 2 of them. Furthermore, we have to paint 7 lines, BUT we can only have 4 lines of any color, so we would need 4 of one color and 3 of another. So, we're going to have duplicate entries; we just have to account for all of them.

With red, blue and yellow, we can have these combos:
red and blue
red and yellow
blue and yellow

Let's use red and blue as an example (and then we'll triple the result to cover all of the combos)

We'll have either
4 red and 3 blue
4 blue and 3 red

With 7 lines of paint, this math will require a permutation AND we have to factor out the duplicates:

7! / (4!)(3!) = 35

So we 4 red and 3 blue lines, there would be 35 options
With 4 blue and 3 red lines, there would also be 35 options

Red and Blue would give us a total of 70 options

Now we triple that to account for every combo of colors....

Final Answer:[spoiler]210 (B)[/spoiler]


GMAT assassins aren't born, they're made,
Rich