Dividing out the swaps you cannot see when some objects are identical

0:00Four letters, all different: R, O, S and E. You know this count already - four times three times two times one, twenty-four arrangements, and every one of them looks different from every other. Now change one letter. R, O, O, T. Still four objects, still four places, so the old argument still gives twenty-four ways of putting them down. But there are not twenty-four different words. Build them and count what you can actually see, and the answer is twelve.

0:35Twenty-four ways, twelve results. Something is being counted twice over, and the something is a swap you cannot see: exchanging the two O's. Here is the move that makes the old argument apply again. Pretend the two O's can be told apart. Put a small one on the first and a small two on the second. Now the four objects are genuinely different, and the count you already know is exactly right: twenty-four.

1:08That is a lie, and it is a deliberate one. You have not solved the problem, you have replaced it with one you can already solve. The whole method is a round trip, and the second half of it is taking the labels off again. Notice what a label is not. It is not part of what the object shows. The letter is what you see; the label is there only so that two things which look identical can be told apart while the counting happens.

1:41So take the labels off and see what merges. R, O-one, O-two, T spells ROOT. So does R, O-two, O-one, T. Two labelled strings, one word. Do that to all twenty-four and they fall into rows, one row for each word that gets spelled. Twelve rows. There they are: ROOT, TOOR, ROTO, TORO, RTOO, TROO, OORT, OROT, OTOR, ORTO, OTRO, and OOTR. Twelve words. Now stop looking at the count and look at the rows themselves.

2:15Every single row holds exactly two labelled strings. Not one anywhere, not three anywhere. Two, twelve times over. Twelve rows of two is twenty-four, which is where we came in. That evenness is not a coincidence, and it is the only reason the next step is allowed at all. Fix one word - take OROT. Which labelled strings spell it? The two O's have landed in the first place and the third. Every other letter is already decided. So the only freedom left is which O carries the one and which carries the two.

2:56That is two ways. And nothing in that argument mentioned OROT. Whatever word you pick, the repeated letters sit in some places, and the labelled strings that spell it are exactly the ways of handing the labels out among those places. So how many labelled strings land on a word does not depend on which word. It depends only on how many of the objects are alike. Now the division, and it is worth being slow about what is being divided by what.

3:30On the top, twenty-four: the labelled strings. Underneath, two: how many labelled strings land on each word. The answer, twelve, is the number of words. Say that in words rather than in symbols. The total, divided by how many land on each, gives how many there are. That is not a special rule for letters. That is what division means. Twenty-four things sorted into rows, every row holding two, is twelve rows - and you never counted the rows.

4:05The two came from the two O's. It is the number of ways of shuffling the labels inside a group of two. Everything so far rests on those rows being the same size. So it is worth seeing what happens when they are not. Take four objects, two of one kind and two of another. There are six arrangements you can tell apart. Merge them by a different rule. Put two together when one is the other slid round in a circle.

4:37Nothing is wrong with that rule. It is what you would do if the row were a bracelet rather than a word. But now the blocks come out in sizes four and two. Not equal. And watch what a division does here. There are two blocks. Six divided by two is three. Three is a whole number, and it is the wrong answer. So a whole number is not evidence. The rows being equal is the thing that has to be checked, and the picture is what checked it.

5:10Now more than one repeated group. Take the word INSTITUTE - nine letters, with I twice and T three times. Label them all and there are nine factorial labelled strings, as before. How many land on each visible word? The two I's can be shuffled among their places in two ways. The three T's among theirs in six. Those two choices do not interfere, so each word answers to two times six, which is twelve.

5:42Nine factorial over twelve is thirty thousand two hundred and forty. And here is where the marks go. Divide by two times three instead - by the counts rather than by their factorials - and you get sixty thousand four hundred and eighty. Exactly twice too big. The reason nobody notices is that on a word whose only repeat is a pair, the two divisors agree. Two and two factorial are the same number.

6:15Say it once in general. Some number of objects in a row, with groups of them alike, and anything left over unlike everything else. The count is the factorial of how many objects there are, divided by the product of the factorials of the group sizes. The second mistake lives here. The word SEVENTEEN has four E's and two N's - six repeated letters in total. The divisor is not six factorial. It is four factorial times two factorial, which is forty-eight, and the answer is seven thousand five hundred and sixty.

6:56Six factorial is seven hundred and twenty, and would have given five hundred and four. Each group is shuffled inside itself. The groups are never shuffled into each other. None of that has been taken on trust, and here is where it gets checked. Take every shape a word of up to six letters can have - every way its letters can fall into repeated groups. There are twenty-nine of them.

7:25For each one, build every labelled arrangement, take the labels off, and see what merges. That is the count, and it owes nothing to any formula. Then put three numbers beside each other: the count you built, the count the division gives, and the count the formula gives. Shapes where the built count and the formula disagree: none. Shapes where the built count and the division disagree: none. Shapes where the rows come out in more than one size: none.

7:58And shapes where that one size is not the product of the group factorials: none. None of this is about the alphabet. Nine discs in a row: four blue, three green, two yellow, and two discs of the same colour cannot be told apart. Same argument, same count. Nine factorial on the top, and underneath four factorial times three factorial times two factorial, which is two hundred and eighty-eight. One thousand two hundred and sixty arrangements.

8:35Flags, beads, votes, identical parts coming off a production line - if two of them are interchangeable, the swap between them is one you cannot see, and it has to be divided out. Now stack restrictions on top. Take INDEPENDENCE: twelve letters, with N three times, E four times and D twice. With nothing restricted, that is one million six hundred and sixty-three thousand two hundred arrangements. Pin the P to the front. Eleven letters are left, and they still carry three N's, four E's and two D's: one hundred and thirty-eight thousand six hundred.

9:19Pin the I to the front and the P to the back, and ten letters are left in the middle: twelve thousand six hundred. Now hold the five vowels together instead. Treat the block as a single object. That leaves eight objects, among which three N's and two D's still repeat, giving three thousand three hundred and sixty. And inside the block the five vowels rearrange - but four of them are E's, so that is five ways, not a hundred and twenty.

9:52Three thousand three hundred and sixty times five is sixteen thousand eight hundred. The division and the block are not two separate tricks; here they are two lines of the same calculation. One more restriction, and this one counts positions rather than letters. PERMUTATIONS has twelve letters, and only one of them repeats: T, twice. Pin the P to the front and the S to the back, and ten letters are left in the middle, still carrying the two T's. One million eight hundred and fourteen thousand four hundred.

10:32Hold the five vowels together instead, and the block plus seven consonants makes eight objects with T still twice - twenty thousand one hundred and sixty. This time the five inside the block are all different, so that really is a hundred and twenty ways in there, and the product is two million four hundred and nineteen thousand two hundred. Now a harder one. How many arrangements have exactly four letters standing between the P and the S?

11:04That is a question about places, not about letters. So count the places first. Go through the twelve positions and find the pairs five apart. Counting both orders - P first or S first - there are fourteen of them. For each of those fourteen, the other ten letters fill the remaining places in one million eight hundred and fourteen thousand four hundred ways. Fourteen times that is twenty-five million four hundred and one thousand six hundred.

11:40One more, and it is a trap in the wording rather than in the arithmetic. Take five objects: three of one kind and two of another. There are ten arrangements. Three of them have all three alike ones standing in one unbroken run. So the arrangements in which they are not all together number seven. It is very easy to read that as no two of them touching. It is not that.

12:07Only one of the ten has no two touching. Seven against one, and the two questions differ by six of the ten arrangements. The seven include the ones where two sit side by side and the third stands apart. Those do not have all three together, so they count. Scale it up. MISSISSIPPI has eleven letters, four I's, four S's and two P's, giving thirty-four thousand six hundred and fifty arrangements.

12:39Fuse the four I's into one block and eight hundred and forty of them have all four together. So thirty-three thousand eight hundred and ten do not. So, the method, in one breath. Pretend the identical objects can be told apart, and count as if they were all different. That count is too big, and you know exactly how much too big. Every arrangement you can actually see has been counted once for each way of shuffling the labels inside each repeated group.

13:15So divide by that number - a product of factorials, one for each group. Never a sum, and never a plain count of the repeats. And remember why the division was available at all. Every arrangement was counted the same number of times. When that stops being true, no single division is right, whatever whole number it happens to produce.