Count the outcomes of several choices made in a row by multiplying the number of options at each stage, instead of listing them all.
Last stop you got a probability by throwing hoops, because nothing about a hoop could be reasoned out. Plenty of chances at this fair can be reasoned out, and to do it you need to know how many outcomes there are. Listing them works until it does not. The food stall is doing a lunch deal: pick one of 3 breads, one of 5 fillings and one of 4 drinks. Writing out every possible lunch would take a page, and the answer is one line of arithmetic.
The grid shows the rule. Each of the 3 breads can take any of the 5 fillings, so there are 3 × 5 = 15 sandwiches. Now each of those 15 sandwiches can take any of the 4 drinks, so there are 15 × 4 = 60 lunches. That is the product rule: when a choice is made in stages, multiply the number of options at each stage. It works because each option at one stage opens up the whole set at the next. It runs backwards too: divide the total by the stages you know to find a missing one.
Sometimes an option cannot be used twice. Three prizes are going to three different volunteers out of the 8 running the stalls. The first prize can go to any of the 8. Once it has gone, only 7 people are left for the second, and 6 for the third. So the count is 8 × 7 × 6 = 336. Each stage has one option fewer than the last, because whoever was picked is now out.
The count you work out is often the bottom of a probability. The raffle prints tickets as a letter from A to E followed by two digits, so there are 5 × 10 × 10 = 500 different codes, one ticket each. How many start with A? The letter is fixed and the digits are free: 1 × 10 × 10 = 100. So the chance of drawing a ticket starting with A is 100 / 500 = 0.2.
Two things have to be true before you multiply. The stages must be separate choices, so that every combination really is on offer. If a drink comes with only one bread, count that drink on its own and add it to the rest. And for the count to be a denominator, the outcomes have to be equally likely. The 500 codes divide the chance evenly only because each is printed on one ticket.
When a choice is made in stages, the number of outcomes is the options at each stage multiplied together. If something cannot be picked twice, each stage has one option fewer than the one before it. The count is often the denominator of a probability: count the outcomes you want, count all of them, and divide. Multiplying is only right when the stages are separate choices and every combination is genuinely available.
The hoop stall needed throwing to work out its chance. The food stall does not: its lunch deal is one of 3 breads, one of 5 fillings and one of 4 drinks, and every combination is on the menu. Leo has started writing them all out and is on his second sheet of paper.
Before the drinks, how many bread and filling pairs are there?
Jobs that lean on this skill. Follow one to see everything it is built on.
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.