Factorial and nPr

We introduce a new notation, which will be useful later, when we discuss binomial experiments. Consider a natural number, say 66. The factorial of 66, written

6!6!

is the product of the first 66 natural numbers, that is

6!=6⋅5⋅4⋅3⋅2⋅16!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1

More generally, we have the following:

Definition 1

Consider a natural number nn. The product of the first nn natural numbers

n⋅(n−1)⋅(n−2)⋅...⋅2⋅1n\cdot (n-1)\cdot (n-2) \cdot ... \cdot 2 \cdot 1

is called the factorial of nn, written

n!n!

In particular, we have

1!=12!=2⋅1=23!=3⋅2⋅1=64!=4⋅3⋅2⋅1=245!=5⋅4⋅3⋅2⋅1=120...\begin{array}{lll} 1! &=& 1\\ 2! &=& 2\cdot 1 &=& 2\\ 3! &=& 3\cdot 2\cdot 1 &=& 6\\ 4! &=& 4\cdot 3\cdot 2\cdot 1 &=& 24\\ 5! &=& 5\cdot 4\cdot 3\cdot 2\cdot 1 &=& 120\\ ... \end{array}

The factorial of 00 is defined as

0!=10!=1

which seems to be totally arbitrary at the moment, but it turns out to be a wise choice. See later.

Warning

Note that !! binds stronger than ++ and ⋅\cdot. So

10+4!=10+4⋅3⋅2⋅1=10+2410+4! = 10+ 4\cdot 3\cdot 2\cdot 1=10+24

and

10⋅4!=10⋅4⋅3⋅2⋅110\cdot4! = 10\cdot4\cdot 3\cdot 2\cdot 1

In particular, 10+4!≠14!10+4!\neq 14! and 10⋅4!≠40!10\cdot4!\neq 40!

You can also also find the factorial on the calculator.

Exercise 1

Solve without calculator:

  1. (4−2)!= ?(4-2)!=\, ?

  2. 4⋅4!= ?4\cdot 4!=\,?

  3. 2!⋅5!= ?2! \cdot 5!=\,?

  4. 10!6!= ?\frac{10!}{6!}=\,?

  5. 1000!998!= ?\frac{1000!}{998!}=\,?

  6. It is n!=120⋅119!n!=120\cdot 119!. Determine the value of nn.

Solution
  1. (4−2)!=2!=2⋅1=2‾(4-2)!=2!=2\cdot 1 =\underline{2}
  2. 4⋅4!=4⋅4⋅3⋅2⋅1=96‾4\cdot 4! = 4\cdot 4\cdot 3\cdot 2\cdot 1 = \underline{96}
  3. 2!⋅5!=2⋅1⋅5⋅4⋅3⋅2⋅1=240‾2! \cdot 5! = 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1= \underline{240}
  4. 10!6!=10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1=10⋅9⋅8⋅7=5040‾\frac{10!}{6!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 }{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 }=10\cdot 9\cdot 8\cdot 7=\underline{5040}
  5. 1000!998!=1000⋅999⋅998⋅...⋅1998⋅...⋅1=1000⋅999=999 000‾\frac{1000!}{998!}=\frac{1000\cdot 999\cdot 998 \cdot ...\cdot 1}{998\cdot ...\cdot 1}=1000\cdot 999=\underline{999\,000}
  6. n!=120⋅119!=120⋅119⋅118⋅...⋅2⋅1=120!→n=120‾n!=120\cdot 119!=120\cdot 119\cdot 118\cdot ...\cdot 2\cdot 1 = 120! \rightarrow n=\underline{120}

6!6! is the multiplication of the first 66 natural numbers. We can generalise the factorial as follows: We write 6P46P4 (say "66 p 44") if we multiply only the 44 highest numbers starting at 66, that is

6P4=6⋅5⋅4⋅3⏟4 numbers=3606P4 =\underbrace{6\cdot 5\cdot 4\cdot 3}_{4 \text{ numbers}}=360

Note that we can express this number with factorials:

6P4=6⋅5⋅4⋅3=6⋅5⋅4⋅3⋅2⋅12⋅1=6!2!=6!(6−4)!6P4 =6\cdot 5\cdot 4\cdot 3=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1}=\frac{6!}{2!}=\frac{6!}{(6-4)!}

Similar, we have

6P1=6⏟1 number=6!(6−1)!6P2=6⋅5⏟2 numbers=6!(6−2)!6P3=6⋅5⋅4⏟3 numbers=6!(6−3)!6P5=6⋅5⋅4⋅3⋅2⏟5 numbers=6!(6−5)!6P6=6⋅5⋅4⋅3⋅2⋅1⏟6 numbers=6!\begin{array}{llll} 6P1 &=&\underbrace{6}_{1 \text{ number}}&=& \frac{6!}{(6-1)!}\\ 6P2&=&\underbrace{6\cdot 5}_{2 \text{ numbers}}&=& \frac{6!}{(6-2)!}\\ 6P3 &= &\underbrace{6\cdot 5\cdot 4}_{3 \text{ numbers}}&=& \frac{6!}{(6-3)!}\\ 6P5 &=&\underbrace{6\cdot 5\cdot 4\cdot 3\cdot 2}_{5 \text{ numbers}}&=& \frac{6!}{(6-5)!}\\ 6P6 &=&\underbrace{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}_{6 \text{ numbers}}=6!\\ \end{array}

Let's summarise:

Theorem 1

Consider a natural number nn. The multiplication of the first rr highest numbers, starting at nn:

n(n−1)...(n−r+1)⏟r numbers\underbrace{n(n-1)...(n-r+1)}_{r \text{ numbers}}

is denoted by

nPrnPr

It is

nPr=n!(n−r)!nPr = \frac{n!}{(n-r)!}

In particular,

nPn=n!nP1=n\begin{array}{lll} nPn&=&n!\\ nP1&=&n \end{array}

Note that nPrnPr can also be found on your calculator.

Exercise 2

Without calculator, determine

  1. 5P25P2
  2. 7P47P4

Then verify your results using the calculator.

Solution
  1. 5P2=5⋅4=20‾5P2=5\cdot 4=\underline{20}
  2. 7P4=7⋅6⋅5⋅4=840‾7P4=7\cdot 6\cdot 5\cdot 4=\underline{840}