05.06.2019: Permutations and Combinations, 4/5: Combinations

Today's soundtrack is Strauss: Masters of Classical Music, Vol. 4.

This afternoon, I'm learning about combinations. Combinations are like permutations, except in a combination, you're looking for a group of things, and their order does not matter.

If I am making a pizza, and I have five toppings to choose from, but I want to put only three of them on, how many ways could I arrange the toppings from bottom to top? Would the cheese go on the bottom or the top? Would the olives be in the middle or on the bottom? This problem is a permutation, because order matters. A permutation will take the form n! / (n - r)!, or nPr (n permute r).

If I am making a pizza, and I have five toppings to choose from, but I want to only put three of them on, and I don't care to know about their order, how many combinations of toppings could I have? Would I get cheese with olives and pickles? Would I get olives with cheese and salami? This problem is a combination, because order does not matter. A combination will take the form n!/r!(n - r)!, or nCr (n choose r). Since the order does not matter, a combination will have less possible results than a permutation will.

Sometimes we will want to choose more than one combination from a large set. For example, if I am running a logistics company and I want to start up a fleet of delivery vehicles, I could go to a dealership and see what they can offer me. If they have seven trucks (each a different model) and five vans (each a different colour), and I want to buy three trucks and two vans, in how many ways can I select the vehicles for my fleet? Well, we can work with two sets of nCr multiplied together. (7C3)(5C2) = (35)(10) = 350 possible combinations of vehicles for my fleet. Notice that the number of vehicles I had to choose from totaled twelve, and the sum of n in my two equations of nPr was 12. Further, the number of vehicles that I wanted was five; the sum of r in my equations was five. Ensure these values match to avoid errors!

As with nPr, we can calculate nCr with the TI-83 Plus in the "Math" function area.

If we are asked to find a combination with "at least x" or "at most x," we are being asked to add together all the different possibilities. For example, if I have ten M&Ms, five of which are orange, and five of which are yellow, and I want to find how many different ways I could arrange five M&Ms with at least three yellow M&Ms in the mix, I will recognize that this gives me four possible scenarios: one in which I have no yellow M&Ms and five orange M&Ms, a second in which I have one yellow M&M and four orange M&Ms, a third in which I have two yellow M&Ms and three orange M&Ms, and a fourth in which I have three yellow M&Ms and two orange M&Ms. So I would calculate (5C0)(5C5) + (5C1)(5C4) + (5C2)(5C3) + (5C3+5C2).

That's it for today! Next time, I'll be learning about Pascal's Triangle!