Rachel has ribbons that are blue, green, red, orange, and ye

I really like questions like this because they can bait you into doing more math than necessary. The combinatorics formula ( \(\frac{n!}{k!(n-k)!}\)) will yield the right answer here if you apply it to both the 1-ribbon( \(\frac{5!}{4!1!}\)) and 2-ribbon possibilities (\(\frac{5!}{3!2!}\)). The first formula will yield 5, and the second one will end up giving 10, so that's 15 total possibilities. Answer choice C.

You can get the answer pretty quickly using the formula, but if you're not comfortable with formulas, feel free to write out the possibilities logically. Any of the five colors works for the 1-ribbon strategy, so the only question is for the 2-colors. If I start with blue, I can add a green, red, orange, or yellow ribbon, and get 4 possibilities. I've now done everything I can with blue, because blue-blue isn't allowed and blue-green and green-blue is the same thing (especially if a gust of wind turns the plant around!). So we're done with blue, on to green! I can add red, orange or yellow to green, so that's 3 more choices. Now green is done, so the next choices are red-orange or red-yellow for 2 more options. The final choice is orange-yellow, so 1 more. The total is 4+3+2+1, which is a pretty obvious pattern when you write them out. Adding back the five monochromatic options, and the final answer is 15, or C).

The explanation for writing out choices may have been longer than the combinatory formula, but it's often much more intuitive and less error-prone. Feel free to write out the options if it's a reasonable amount, or extrapolate the choices logically if it's not. The formula is very helpful here, but not at all necessary for solving the question.

Hope this helps!
-Ron