Factorials and Counting Notation

Why Factorials Appear in Counting

When a sequence of choices is made, the number of possibilities can be counted by multiplying the number of choices at each stage. For example, if five distinct objects are placed in five ordered positions, there are five choices for the first position, four remaining choices for the second, then three, two, and one. The product \(5\cdot4\cdot3\cdot2\cdot1\) is easy to write, but longer descending products are cumbersome. Factorial notation gives these products a compact form.

This notation is useful both as arithmetic shorthand and as a way to express counts. It does not change the counting principle: each factor records how many choices remain at a stage. The objects in an arrangement must be distinguishable for those choices to count different arrangements.

Evaluating Factorials

To evaluate a factorial, write every factor in the descending product down to \(1\), then multiply. The exclamation mark is part of the notation: \(6!\) means “six factorial,” not “six” followed by punctuation. Factorials grow quickly because each new value multiplies the previous product by another integer.

The recurrence follows by separating the first factor from the defining product. For example, the factors of \(n!\) are \(n\), followed by \((n-1)(n-2)\cdots1\). The latter product is exactly \((n-1)!\), so \(n!=n(n-1)!\). This identity is useful for evaluating factorials and for simplifying expressions.

Zero Factorial and the Empty Product

The definition \(0!=1\) can seem surprising because there are no positive integers from \(0\) down to \(1\) to multiply. The value \(1\) is chosen for a mathematical reason: it keeps the factorial recurrence consistent at its smallest step. Substituting \(n=1\) into \(n!=n(n-1)!\) gives \(1!=1\cdot0!\). Since \(1!=1\), this requires \(0!=1\).

There is also a counting interpretation. If there are no objects to arrange, there is exactly one arrangement: the empty arrangement, in which nothing is placed. Thus the number of arrangements of zero distinct objects is \(1\), matching \(0!=1\). This convention makes formulas and counting arguments work smoothly at the boundary case.

Simplifying Factorial Expressions

A factorial may contain many factors that do not need to be calculated individually. When a factorial appears in a numerator and a smaller factorial appears in a denominator, expand the larger factorial only far enough to expose the smaller factorial as a common product. Then cancel that common nonzero factor. This avoids calculating very large values and reduces arithmetic errors.

For example, the recurrence gives \(8!=8\cdot7!\). Therefore \(8!/7!=8\), without needing to evaluate either factorial as a large number. More generally, when \(n\) and \(k\) are nonnegative integers with \(n\geq k\), repeated use of the recurrence yields $$ \frac{n!}{k!}=(k+1)(k+2)\cdots n \quad\text{if } n>k, $$ and the ratio is \(1\) when \(n=k\). The denominator \(k!\) is always positive, so this cancellation is valid.

The order in which factors are written does not affect their product, but writing factorial expansions in descending order makes the cancellation visible. A useful habit is to leave a factorial unexpanded until you know which factors are needed. If the numerator and denominator are both evaluated first, the numbers can become unnecessarily large.

Factorials as Counts of Full Arrangements

Consider \(n\) distinct objects to be placed in \(n\) ordered positions, using each object exactly once. There are \(n\) choices for the first position. Once that object is placed, \(n-1\) remain for the second position; then \(n-2\) remain for the third. Continuing in this way, the last two positions have two choices and then one choice. By the multiplication principle for sequential choices, the total number of arrangements is the product of these stage counts.

This is a direct use of factorial notation to compress the product of sequential choice counts. It concerns arrangements of all the objects. Counting arrangements that use only some of a larger collection is a different question and requires keeping track of how many positions are filled.

Common Mistakes and Why the Notation Matters

• Stopping the product too soon. Writing \(5!=5\cdot4\) leaves out the factors \(3\), \(2\), and \(1\). Expand all the way down to \(1\) when evaluating a factorial.
• Treating \(0!\) as zero. By definition \(0!=1\), not \(0\). This value agrees with the recurrence and with the single empty arrangement.
• Canceling part of a factorial incorrectly. In \(\frac{9!}{6!}\), canceling \(6!\) leaves \(9\cdot8\cdot7\), not just \(9\). First expand the larger factorial down to the smaller one.
• Using a full-arrangement count when not all objects are used. The count \(n!\) applies when all \(n\) distinct objects occupy \(n\) ordered positions. It does not by itself describe every selection or partial arrangement.
• Ignoring whether objects are distinct. The multiplication \(n(n-1)\cdots1\) counts separate choices at each stage. If some objects are indistinguishable, different sequences of choices may not correspond to different visible arrangements.

Factorials are more than a compact way to write multiplication. They let us recognize the same descending product in different forms, simplify ratios without producing huge numbers, and express the number of full arrangements of distinct objects. Writing the factors or stating the counting assumptions makes the meaning of a factorial expression clear before using it.