6th Edition: Guide:Numbered Properties: Pg 62: "The yearboo"

[ShayaanN905]

Hi MPrep,

I had a question regarding a problem at the end of Chapter 4: "Combinatorics" of the 6th Edition, "Number Properties" Strategy Guide. The question appears on page 62, and is as follows:

"The yearbook committee has to pick a color scheme for this year's yearbook. There are 7 colors to choose from (red, orange, yellow, green, blue, indigo, and violet). How many different color schemes are possible if the committee can select at most 2 colors?"

The answer to the question seems to use the anagram approach, and boils the answer down to adding the combinations possible if 1 color is chosen and if 2 colors are chosen. While I understand the method used to calculate the combinations for 1 color being chosen, 7! / 1! 6!, I'm having trouble comprehending the method used to calculate the combinations for 2 colors chosen.

From my understanding, the anagram method is used when you cannot distinguish between two options, because they are "identical", such as the example at the top of page 62 with the I Eta Pi fraternity choosing between "identical" members, or the example earlier in the chapter with giving out two identical "silver" medals to racers (pg 60).

In this problem, my understanding is that each color is distinguishable from one another (there is only one orange, one blue, one green, etc), and therefore, the method for calculating the combinations for 2 colors could not equal 7! / 2! 5!.

I'm having trouble understanding where the 2! comes from here, as I believe each color is distinguishable from one another when being picked. The answer in the book seems to contradict the language of "identical" in the previous examples (specifically the example directly above it).

I would really appreciate any clarity you could provide, especially with my GMAT exam date around the corner!

Regards,

Shayaan

[Sage Pearce-Higgins]

The anagram approach is a useful strategy, but it's important that it doesn't replace logic and common sense. After all, in a problem like this one, in which applying the anagram approach is not immediately obvious, you want to have a backup strategy.

If I say 'I have 7 colors; how many individual single colors can you choose?' we can see that there are 7 possibilities without recourse to the anagram method. If I'm picking 2 colors from the set of 7, then we can think 'I choose a first color, that's 7 possibilities, then I have 6 colors to choose from for the second choice. So, for each of the first choice colors, I have 6 more possibilities. That makes 7 x 6 = 42 possibilities in total.' If that reasoning isn't clear to you, then start to list out the combinations, and you'll see the pattern. You mention 'identical' here, and that's important because, in this situation order doesn't matter. A first-choice green and second-choice blue gives the same color combination as first-choice blue and second-choice green. This means that, in my 42 possibilities, I've counted each one twice and the real total is 21.

The above relies more on logic and seeing patterns. I'd encourage you to get familiar with the idea of order mattering or not, and with the concept of first-choice, second-choice etc.

Applying the anagram method to this problem would work as follows. Consider the number of choices in which 2 colors are chosen and list out the colors in the top row, and then 'chosen' or 'not' in the bottom row:
R O Y G B I V
C C N N N N N

That results in the factorial expression given on page 62.