Suppose we want to choose 2 objects, without replacement, from the 4 objects pencil, eraser, desk, and chair. (a) How many ways can this be done, if the order of the choices matters? (b) How many ways can this be done, if the order of the choices does not matter? \( \square \)

When order matters, you can think of it as a fun game where you first pick an object, then pick another one from the remaining choices. For the first pick, you have 4 options, and for the second pick (since you don’t put the first object back), you have 3 options. Thus, the total number of arrangements is \(4 \times 3 = 12\) ways! Now, if order doesn’t matter, it’s like making a combo selection where it’s all about the items you end up with, not the sequence. You need to choose 2 out of the 4, which can be done using combinations. The formula is \(\binom{n}{r}\) which translates here to \(\binom{4}{2} = 6\) different groups of objects!