Why choices made one after another multiply rather than add

0:00Here is a lock with four wheels, and each wheel carries the ten digits. It opens on one particular ordered sequence, with no digit used twice. You remember the first digit. It is seven. The other three are gone. So how many sequences would you have to try? Listing them is not an option, and you are about to be able to answer that without listing anything. The answer is five hundred and four, and the whole of this is about where that number comes from.

0:34Because it does not come from counting a list. It comes from an argument, and the argument is short. Start somewhere small enough to check by hand. You have three pairs of trousers and two shirts, and you are going to wear one of each. How many different outfits is that? A very natural first answer is five, and five is the number of garments in the wardrobe. It is not the number of outfits, because an outfit is not a garment. It is a pair.

1:08Build them instead of counting them. Trousers one with shirt one. Trousers one with shirt two. Trousers two with shirt one, trousers two with shirt two. Trousers three with shirt one, trousers three with shirt two. Six. Not five. And nothing about that list was multiplied. It was made, and then counted. Now look at the shape of what was just built, because the shape is the whole argument. The six outfits fall into three groups, one for each pair of trousers, and every group holds two.

1:46Two and two and two. Add those up and you get six. That is not a coincidence and it is not a second method. Multiplying is what adding equal groups is called. Three times two means three groups of two, and the only reason six is the answer is that the groups really are equal. So the question to ask is never whether to multiply. It is whether the groups come out equal.

2:15And here is what makes them equal: whichever trousers you pick first, you are left with the same number of shirts. That sentence is the counting principle. Everything else is applying it. Now be careful about what has to be the same, because it is easy to ask for too much. In the trousers and shirts, every first choice leaves you the very same two shirts. Nothing was used up.

2:44That is the comfortable case, and it is not the general one. Take the four letters R, O, S, E, and four places to put them in, using each letter once. Pick R first and you are left with O, S and E. Pick O first and you are left with R, S and E. Those are different collections. There is no shared pile of leftovers here at all. But every one of the four first choices leaves exactly three letters, and three is all the argument needs.

3:18The contents may change from branch to branch. Only the count has to hold still. Which raises the obvious question. What happens when the counts do not hold still? Go back to the three pairs of trousers and two shirts, and change one thing. One shirt clashes with one pair. Build the outfits again and there are five of them, not six. The second stage now leaves two open, then two, then one. Those are not equal groups, so there is nothing to multiply by.

3:54Three times two would have said six, and six is one too many. Here is a second one, further from the edge case. Pick a number from one, two, three, and then pick a strictly smaller one. After one there is nothing smaller to pick. After two there is one thing. After three there are two. Nought, one and two. There is no single number that the second stage is worth, so the product is not merely wrong. It cannot even be written down.

4:27Build them and there are three: two then one, three then one, three then two. Nothing so far needed there to be only two stages. Two school bags, three lunch boxes, two water bottles, one of each. Take the bags and boxes first. Every bag leaves the same three boxes, so that is three groups of two, or two groups of three: six pairs. Now hang the bottles off those six. Every one of the six pairs leaves the same two bottles.

5:02Six groups of two. Twelve. And notice what the third stage did. It doubled what was already there, because the number it left open was two. A stage does not add itself to the count. It multiplies what is already built by however much it leaves open. That is why a modest number of stages runs away from you so quickly. Back to the four letters and the four places, and let us actually do it.

5:33The first place can take any of the four letters. Whatever went in there, the second place has three letters left. Not the same three each time, but three each time. The third place has two, and the last place has the one letter nobody used. Four groups, each containing three groups, each containing two, each containing one. Twenty-four. The factors shrink because the supply shrinks. Each place uses a letter up, and the next place is one poorer.

6:07That is not a law of counting. It is a fact about this particular rule, that a letter may not be used twice. Change the rule and the factors stop shrinking. So change it. Same four places, same four letters, but now a letter may be used as often as you like. The first place takes four. The second place takes four, because nothing was used up. So does the third, and so does the fourth. Two hundred and fifty-six.

6:40Twenty-four and two hundred and fifty-six, on the same four places, from the same four letters. Only the rule about reuse changed. It is tempting to think of the no-reuse rule as throwing away some tidy share of the arrangements. It does not. Divide two hundred and fifty-six by twenty-four and you get thirty-two thirds, which is not a whole number. The restriction is not a fraction taken off the top. It is a different count, arrived at by a different sequence of factors.

7:18There is one more question the principle does not answer for you, and it decides whether you can use the principle at all. Which stage do you do first? Make a two-digit even number from the digits one to five. Even means the units digit has to be two or four. Fill the units place first. Two ways. Then the tens place: five ways, whatever the units digit was. Ten numbers. And if you fill the tens place first instead, five ways, then two ways, and it is ten again.

7:53With repeats allowed the order is a convenience. Now bar the repeats and try both orders again. Units first: two ways, and then four ways every single time, because whichever even digit you used, four of the five remain. Two groups of four. Eight. Tens first: five ways. Then how many even digits are left? If you led with two, only one. If you led with one, both. Two, one, two, one, two. Not equal, so there is nothing to multiply by, and the principle simply does not apply in that order.

8:34The answer is still eight either way. But only one of the two orders can be counted this way. So the rule of thumb has a reason behind it. Fill the restricted place first, and the restriction is spent before it can make the groups uneven. Everything so far has multiplied. Here is a problem where you have to do both. Five flags, all different colours, stacked on a staff to make a signal. A signal uses at least two of them, in order.

9:08So a signal might be two flags, or three, or four, or all five. Count each length on its own. Two flags: five ways then four ways. Twenty. Three flags: five, four, three. Sixty. Four flags: five, four, three, two. A hundred and twenty. Five flags: five, four, three, two, one. A hundred and twenty again, and it is the same number, because the last stage leaves exactly one option and multiplying by one changes nothing.

9:43Now, the total. Twenty and sixty and a hundred and twenty and a hundred and twenty. Three hundred and twenty. Added, not multiplied. And the reason is not that we ran out of things to multiply. The reason is that those four are not stages. They are alternatives. A stage is something you pass through. You choose trousers and then you also choose a shirt. An alternative is something you choose between. A signal is two flags long or it is three, and it is never both.

10:20Stages multiply. Alternatives add. And alternatives may only be added when no outcome belongs to two of them. That condition is doing real work. Watch what happens without it. Split the same signals a different way: the ones using at least three flags, and the ones using at most three. Between them they cover everything, and both counts are honest. Three hundred, and eighty. Add those and you get three hundred and eighty, against a true total of three hundred and twenty.

10:56Sixty too many, which is exactly the number of three-flag signals, because every one of them was counted in both halves. Two correct numbers, added correctly, giving a wrong answer. Nothing about the arithmetic was wrong. The split was. So, the lock. Four wheels, ten digits each, no digit twice, and the first one is known to be seven. That leaves three wheels to fill, and nine digits to fill them from, since seven is already spent.

11:31The first unknown wheel takes any of nine. Whatever went there, the second takes eight, and the third takes seven. Nine groups of eight groups of seven. Five hundred and four. Nobody drew that. Nobody listed it. And no branching picture was needed, because the argument never depended on being able to see the whole thing. It only ever needed one sentence: every choice at one stage leaves the same number open at the next.

12:04That sentence is what you check. If it holds, multiply. If the outcomes come in kinds that cannot overlap, add. And if the number a stage leaves open depends on what came before it, do not reach for either. Reorder the stages, or split into cases, until it does not.