If we use (as mathematicians sometimes do) the notation \(T^S\) to denote the set of all functions from \(S\) to \(T\) and the notation \(|{X}|\) to denote the number of elements in a set \(X\), then the equation

(*) \(|{T^S}| = |{T}|^{|{S}|}\)

(where the superscripting on left side of the \(=\) sign is the notation I’ve just described for sets of functions and the superscripting on the right side of the \(=\) sign denotes ordinary exponentiation) applies in all cases, including cases where \(S\), \(T\), or both are empty, provided that we take \(0^0\) to be \(1\). Let’s take a closer look at how this works, first when \(|{S}|\) and \(|{T}|\) are nonempty, and then when \(S\), \(T\), or both are empty.

• As an example of Equation (*) applying where \(|{S}|\) and \(|{T}|\) are nonempty, suppose we want to count the number of possible ways to distribute a set \(S\) of \(5\) objects among a set \(T\) of \(3\) bins. Then we have \(3\) choices of bin for the first object, \(3\) choices for the second object, etc., giving a total of \(3\times 3\times 3\times 3\times 3 = 3^5 = |{T}|^{|{S}|}\) possibilities altogether.

• If \(S\) is nonempty but \(T\) is empty, then the number of functions from \(S\) to \(T\) is zero. [Proof: Choose some element \(s\in S\). Then for any function \(f\) from \(S\) to \(T\), we must have \(f(s)\in T\). But this is impossible because \(T\) is the empty set.] So we see that \(|{T}^{S}|=|{\emptyset^S}| = 0\), and this is indeed the same as \(|T|^{|{S}|} = |{\emptyset}|^{|{S}|} = 0^{|{S}|} = 0\). That is, the rule \(|{T^S}| = |{T}|^{|{S}|}\) indeed applies in cases where \(S\) is nonempty but \(T\) is empty. Notice, however, that our proof that \(|{\emptyset^S}| = 0\) does *not* apply to the case \(S=\emptyset\) (equivalently, \(|{S}| = 0\)), since it depends on the possibility of choosing some \(s\in S\).

• Finally, if \(S\) is empty, then there is exactly one function from \(S\) to \(T\), namely the empty function, and this is so regardless of whether \(T\) is empty or nonempty. Thus, we are led to the conclusion that it makes sense to say \(|{T}|^0 = |{T}|^{|{\emptyset}|} = |{T}^{\emptyset}| = 1\) for every possible value of \(|{T}|\) including \(0\).

To recap, the preceding analysis shows that the equation \(|{T^S}| = |{T}|^{|{S}|}\) applies for *all* finite sets \(S\) and \(T\), empty or not, provided that we define \(0^0\) as \(1\). It think it also provides some insight into *why* it makes sense, in this context, to extend the rule \(n^0=1\), but *not* the rule \(0^n=0\), from the case where \(n>0\) to the case \(n=0\).

One might say, “Okay, I see why it make sense to say \(0^0=1\) in this context, but how important is it? How often do we care about counting functions from the empty set to the empty set, anyway?” Without going into details, I’ll simply remark that in combinatorics we often make use of *recurrence relations* that give answers to various counting problems as functions of the answers to simpler counting problems, and that it is often convenient to state such recurrences so that the base cases are trivial counting problems involving sets, functions, strings, etc. of size \(0\).

One final, slightly off-topic, comment: Mathematicians have developed a number of different ways to define addition, subtraction, and exponentiation when one or both arguments are infinite. In one of those systems, known as cardinal arithmetic, exponentiation is defined so that Equation (*) applies to *all* sets \(S\) and/or \(T\) whether finite or infinite.

We’ll be happy to.

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]]>I’ve seen a number of people recently claiming that the word “term” (or, more or less equivalently, “monomial”) in itself solves the problem. You seem to be supposing that the word **means** “numbers and variables written together without an explicit operator”, which is what we mean by juxtaposition or implicit multiplication, and that it is inherently treated as a unit. But that is not what it means (in any source I can find). Perhaps you can show me a source that teaches using this word the way you do? I’m always interested in seeing how things are taught in other parts of the world.

A term is commonly defined as one of the entities that are *added or subtracted* in an expression, and which themselves contain only multiplication (sometimes division) and powers, as in a polynomial. (It is used in a couple different ways, sometimes a little loosely, so you may find variations on this, commonly including expressions containing addition inside parentheses.) This definition emphasizes what **operations **are or are not involved, not how they are **written**; and I have not seen it used in explaining the order of operations in your manner. The key idea is that the **additions** in a polynomial are done **last**; divisions are not (at least primarily) in view.

As I’ve said many times, I think it’s *reasonable* to perform these multiplications first; and it would not be unreasonable to use words like “term” to express that idea. If you can provide a source that teaches it that way, I will happily add it to my list. But to my knowledge, that is not a standard way to express the idea, and in any case, your assertion as to what is “always” done does not make it so around the world. I wish it were! But see part 2 of this series (which is where this comment perhaps belongs).

Incidentally, the term can’t have originated *in this blog*, which has only existed since 2018; but it is conceivable that I invented it in early answers on *Ask Dr. Math* in 1998, because it was much harder to find information on such terminology back then (and then it was just a question from students about how to interpret expressions in class, not yet a controversial meme). I don’t recall where I found it.

1÷2a is always interpreted as 1÷(2 x a), and that is because ‘2a’ is a term. If we make a = 3 + b, then we get:

1÷2(3+b), and so ‘2(3+b)’ is also a term, with 2 as the coefficient.

]]>It’s definitely an order of operations issue in that people argue over **which** order of operations is “correct”, and don’t see any ambiguity because they are certain of their interpretation. In particular, I tend to think of it as a dialect question: Some people grow up speaking a dialect of English in which one says, say, “Congress is …”, and others learn a dialect in which “Congress are …”. Both are “correct” in their own cultures. Or, in some cultures a double negative is taken literally (making a positive), while in others, it is merely an intensifier (“it don’t make no difference”), leading two people to hear opposite meanings. Strict PEMDAS is one dialect of math, and IMF is another. Fortunately, most such differences don’t lead to wars.

You are wrong. If 6/2(1+2) is written then you are not following PEMDAS by doing the multiplication first. If, as in PEMDAS, multiplication and division have equal precedence, the rule is to work left to right in written order, so you are wrong. It is an order of operations issue. Following the order of operations results in the answer 9. Doing the multiplication first breaks the order of operations, and produces the answer 1. So it most definitely IS an order of operations issue!

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