Factorial Notation

This section introduces factorial notation, the essential shorthand that underlies every formula developed later in the chapter for permutations and combinations.

Definition: for a natural number $n$, the factorial of $n$, written $n!$ (read 'n factorial'), is the product of the $n$ natural numbers from 1 to $n$: $n! = 1\times2\times3\times\cdots\times(n-2)\times(n-1)\times n$. Equivalently, and often more usefully for computation, it can be written in the reverse order: $n! = n\times(n-1)\times\cdots\times3\times2\times1$. For example, $5! = 5\times4\times3\times2\times1=120$, read as '5 factorial is equal to 120'. A short table of illustrations builds this up from the smallest cases: $1!=1$; $2!=2\times1=2$; $3!=3\times2\times1=6$; $4!=4\times3\times2\times1=24$, and so on, with each new factorial naturally being the previous one multiplied by the next integer.

A crucial convention closes the definition: although 0 is not itself a natural number, we DEFINE $0!=1$. This convention is not arbitrary — it is exactly what is needed to keep later formulas (like ${}^nC_0=1$ and ${}^nP_n=n!/0!=n!$) consistent without a separate special case.

The section then lists eight properties of factorial notation, for positive integers $m,n$: (1) $n! = n\times(n-1)!$; (2) for $n>1$, $n! = n\times(n-1)\times(n-2)!$; (3) for $n>2$, $n! = n\times(n-1)\times(n-2)\times(n-3)!$ — these three properties are simply the same 'peel off the leading factors' idea taken one, two, or three steps deep, and are the basis of the cancellation technique used throughout the chapter to simplify ratios of factorials. (4) $(m+n)!$ is always divisible by BOTH $m!$ and $n!$ — for example, $(3+4)!=7!$ is divisible by both $3!$ and $4!$. Properties (5)–(8) are deliberately-stated WARNINGS against common misconceptions: (5) $(m\times n)! \ne m!\times n!$; (6) $(m+n)! \ne m!+n!$; (7) for $m>n$, $(m-n)! \ne m!-n!$, although $m!$ IS divisible by $n!$; and (8) $(m\div n)! \ne m!\div n!$ — factorials simply do not distribute over addition, subtraction, multiplication, or division the way ordinary exponents or coefficients might, and several exercise problems in this section are built specifically to test whether this distinction has been understood.

Solved Example 1 asks for the value of $6!$: directly, $6!=6\times5\times4\times3\times2\times1=720$.

Solved Example 2 asks to show that $(7-3)!\ne7!-3!$: computing the left side, $(7-3)!=4!=4\times3\times2\times1=24$; computing the right side, $7!=5040$ and $3!=6$, so $7!-3!=5040-6=5034$. Since $24\ne5034$, indeed $(7-3)!\ne7!-3!$, confirming property (7) above with a concrete numerical case.

Solved Example 3 asks to find $n$ if $(n+6)!=56(n+4)!$. Writing $(n+6)!=(n+6)(n+5)(n+4)!$ and dividing both sides by the common $(n+4)!$ gives $(n+6)(n+5)=56$. Rather than expanding this into a quadratic and using the quadratic formula, the solution writes 56 directly as a product of two consecutive integers: $56=8\times7$. Matching the larger factor on each side, $n+6=8$, so $n=8-6=2$. This 'reduce the factorial ratio to a product of consecutive integers, then factor the target number into the same count of consecutive integers' technique is the standard method used repeatedly for 'find n' factorial equations throughout the rest of the chapter, including several exercise problems in this very section. …