\(3!\)
10.5 Factorial notation
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10.4 The fundamental counting principle
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10.6 Application to counting problems
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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:
\(n\times \left(n-1\right)\times \left(n-2\right)\times \cdots \times 3\times 2\times 1\)This allows us to formulate the following:
The total number of possible arrangements of \(n\) different objects is \[n \times (n-1) \times (n-2) \times \ldots \times 3 \times 2 \times 1 = n!\]
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 \(\dfrac{n!}{(n-1)!} = n\)
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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:
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Input the number.
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Press SHIFT on your CASIO or 2ndF on your SHARP calculator.
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Then press \(x!\) on your CASIO or \(n!\) on your SHARP calculator.
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Finally, press equals to calculate the answer.
If we follow these steps for \(12!\), we get the answer \(\text{479 001 600}\).
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- 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:\[\dfrac{n!}{(n-1)!} = \dfrac{n \times \color{red}{(n-1)} \times
\color{blue}{(n-2)}
\times \color{magenta}{(n-3)} \times \ldots \times \color{teal}{3} \times \color{orange}{2} \times
\color{green}{1}}{\color{red}{(n-1)}
\times \color{blue}{(n-2)} \times \color{magenta}{(n-3)} \times \ldots \times \color{teal}{3} \times
\color{orange}{2} \times
\color{green}{1}} = n\]
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:
\(\dfrac{n!}{(n-2)!} = n^{2} - n\)
\(\dfrac{(n-1)!}{n!} = \dfrac{1}{n}\)
\(\dfrac{(n-2)!}{(n-1)!} = \dfrac{1}{n-1} \text{ for } n>1\)
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