\(3!\)
10.5 Factorial notation
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10.6 Application to counting problems

10.5 Factorial notation (EMCK3)
Worked example 12: The arrangement of outcomes without repetition
Eight athletes take part in a \(\text{400}\) \(\text{m}\) race. In how many different ways can all \(\text{8}\) places in the race be arranged?
Any of the \(\text{8}\) athletes can come first in the race. Now there are only \(\text{7}\) athletes left to be second, because an athlete cannot be both second and first in the race. After second place, there are only \(\text{6}\) athletes left for the third place, \(\text{5}\) athletes for the fourth place, \(\text{4}\) athletes for the fifth place, \(\text{3}\) athletes for the sixth place, \(\text{2}\) athletes for the seventh place and \(\text{1}\) athlete for the eighth place. Therefore the number of ways that the athletes can be ordered is as follows: \[8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{40 320}\]
As in the example above, it is a common occurrence in counting problems that the outcome of the first event reduces the number of possible outcomes for the second event by exactly \(\text{1}\), and the outcome of the second event reduces the possible outcomes for the third event by \(\text{1}\) more, etc.
As this sort of problem occurs so frequently, we have a special notation to represent the answer. For an integer, \(n\), the notation \(n!\) (read \(n\) factorial) represents:
\(n\times \left(n1\right)\times \left(n2\right)\times \cdots \times 3\times 2\times 1\)This allows us to formulate the following:
The total number of possible arrangements of \(n\) different objects is \[n \times (n1) \times (n2) \times \ldots \times 3 \times 2 \times 1 = n!\]
with the following definition: 0! = 1.
Worked example 13: Factorial notation
 Determine \(12!\)
 Show that \(\dfrac{8!}{4!} = 8 \times 7 \times 6 \times 5\)
 Show that \(\dfrac{n!}{(n1)!} = n\)

We know from the definition of a factorial that \(12! = 12 \times 11 \times 10 \times \ldots \times 3 \times 2 \times 1\). However, it can be quite tedious to work this out by calculating each multiplication step on paper or typing each step into your calculator. Fortunately, there is a button on your calculator which makes this much easier. To use your calculator to work out the factorial of a number:

Input the number.

Press SHIFT on your CASIO or 2ndF on your SHARP calculator.

Then press \(x!\) on your CASIO or \(n!\) on your SHARP calculator.

Finally, press equals to calculate the answer.
If we follow these steps for \(12!\), we get the answer \(\text{479 001 600}\).

 Expand the factorial notation: \[\dfrac{8!}{4!} = \dfrac{8 \times 7 \times 6 \times 5 \times \color{red}{4} \times \color{blue}{3} \times \color{magenta}{2} \times \color{teal}{1}}{\color{red}{4} \times \color{blue}{3} \times \color{magenta}{2} \times \color{teal}{1}} = 8 \times 7 \times 6 \times 5 = \text{ RHS}\]
 Expand the factorial notation:\[\dfrac{n!}{(n1)!} = \dfrac{n \times \color{red}{(n1)} \times
\color{blue}{(n2)}
\times \color{magenta}{(n3)} \times \ldots \times \color{teal}{3} \times \color{orange}{2} \times
\color{green}{1}}{\color{red}{(n1)}
\times \color{blue}{(n2)} \times \color{magenta}{(n3)} \times \ldots \times \color{teal}{3} \times
\color{orange}{2} \times
\color{green}{1}} = n\]
If \(n = 1\), we get \(\frac{1!}{0!}\). This is a special case. Both \(1!\) and \(0! =1\), therefore \(\frac{1!}{0!} = 1\) so our identity still holds.
Factorial notation
Work out the following without using a calculator:
\(6!\)
\(2!3!\)
\(8!\)
\(\dfrac{6!}{3!}\)
\(6! + 4!  3!\)
\(\dfrac{6!  2!}{2!}\)
\(\dfrac{2! + 3!}{5!}\)
\(\dfrac{2! + 3!  5!}{3!  2!}\)
\((3!)^{3}\)
\(\dfrac{3! \times 4!}{2!}\)
Calculate the following using a calculator:
\(\dfrac{12!}{2!}\)
\(\dfrac{10!}{20!}\)
\(\dfrac{10! + 12!}{5! + 6!}\)
\(5!(2! + 3!)\)
\((4!)^{2}(3!)^{2}\)
Show that the following is true:
\(\dfrac{n!}{(n2)!} = n^{2}  n\)
\(\dfrac{(n1)!}{n!} = \dfrac{1}{n}\)
\(\dfrac{(n2)!}{(n1)!} = \dfrac{1}{n1} \text{ for } n>1\)
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10.4 The fundamental counting principle

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10.6 Application to counting problems
