As a quality control procedure, a chemical lab includes in each sample to be tested a randomly chosen blank. Each blank consists of a single known chemical or a combination of two known chemicals. If protocol requires the lab to have available 12 different blanks, what is the minimum number of chemicals required to prepare the blanks?

To determine the minimum number of chemicals required to prepare 12 different blanks in a chemical lab, consider that each blank can consist of either a single chemical or a combination of two different chemicals. The combinations from a set of "n" chemicals can be determined by the formula for combinations: C(n, k) = n! / (k!(n-k)!), where k is the number of items to choose (in this case, 2 for combinations). For single chemicals, there are "n" possibilities. For combinations of two different chemicals, the number is given by C(n, 2) = n(n-1)/2. So the total number of different blanks you can create is n + n(n-1)/2. Setting this equal to 12 allows you to solve for the minimum "n." Through trial and error or a systematic approach, find that with 5 chemicals, you can create 5 (for the single chemicals) + 10 (combinations of 2) = 15 different blanks, exceeding your goal. However, with only 4 chemicals, you generate 4 (single) + 6 (combinations) = 10, which is insufficient. Therefore, the minimum number of chemicals needed is 5. Isn't it fascinating how combinations multiply your options? It turns a few ingredients into a feast of choices!