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Factorial Notation

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

  1. Determine \(12!\)
  2. Show that \(\dfrac{8!}{4!} = 8 \times 7 \times 6 \times 5\)
  3. Show that \(\dfrac{n!}{(n-1)!} = n\)

  1. 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 \(\text{479 001 600}\).

  2. Expand the factorial notation: \[\dfrac{8!}{4!} = \dfrac{8 \times 7 \times 6 \times 5 \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}} = 8 \times 7 \times 6 \times 5 = \text{ RHS}\]
  3. Expand the factorial notation:\[\dfrac{n!}{(n-1)!} = \dfrac{n \times \cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = n\]

    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

Exercise 10.5

Work out the following without using a calculator:

\(3!\)

\[3 \times 2 \times 1 = \text{6}\]

\(6!\)

\[6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{720}\]

\(2!3!\)

\[2 \times 1 \times 3 \times 2 \times 1 = \text{12}\]

\(8!\)

\[8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \text{40 320}\]

\(\dfrac{6!}{3!}\)

\[\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = \text{120}\]

\(6! + 4! - 3!\)

\[(6 \times 5 \times 4 \times 3 \times 2 \times 1) + (4 \times 3 \times 2 \times 1) - (3 \times 2 \times 1)= 720 + 24 -6 = \text{738}\]

\(\dfrac{6! - 2!}{2!}\)

\[\dfrac{(6 \times 5 \times 4 \times 3 \times 2 \times 1) - (2 \times 1)}{2 \times 1}= \dfrac{720 - 2}{2} = \text{359}\]

\(\dfrac{2! + 3!}{5!}\)

\[\dfrac{(2 \times 1) + (3 \times 2 \times 1)}{5 \times 4 \times 3 \times 2 \times 1}= \dfrac{2 + 6}{120} = \dfrac{1}{15}\]

\(\dfrac{2! + 3! - 5!}{3! - 2!}\)

\[\dfrac{2 + 6 - 120}{6 - 2}= \dfrac{-\text{112}}{4} = -\text{28}\]

\((3!)^{3}\)

\(6 \times 6 \times 6 = \text{216}\)

\(\dfrac{3! \times 4!}{2!}\)

\[\dfrac{(3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1)}{2 \times 1} = \text{72}\]

Calculate the following using a calculator:

\(\dfrac{12!}{2!}\)

\[\dfrac{(12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{2 \times 1} = \text{239 500 800}\]

\(\dfrac{10!}{20!}\)

\(\text{1,49} \times \text{10}^{-\text{12}}\)

\(\dfrac{10! + 12!}{5! + 6!}\)

\(\text{574 560}\)

\(5!(2! + 3!)\)

\(\text{960}\)

\((4!)^{2}(3!)^{2}\)

\(\text{20 736}\)

Show that the following is true:

\(\dfrac{n!}{(n-2)!} = n^{2} - n\)

\begin{align*} \dfrac{n!}{(n-2)!} & = \dfrac{n \times (n-1) \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} \\ \\ &= n(n-1) = n^{2} - n \end{align*}

\(\dfrac{(n-1)!}{n!} = \dfrac{1}{n}\)

\[\dfrac{(n-1)!}{n!} = \dfrac{\cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{n \times \cancel{(n-1)} \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = \dfrac{1}{n}\]

\(\dfrac{(n-2)!}{(n-1)!} = \dfrac{1}{n-1} \text{ for } n>1\)

\[\dfrac{(n-2)!}{(n-1)!} = \dfrac{\cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}}{(n-1) \times \cancel{(n-2)} \times \cancel{(n-3)} \times \ldots \times \cancel{3} \times \cancel{2} \times \cancel{1}} = \dfrac{1}{n-1}\]