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Multiplication Rule |
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If an event can occur in \(m\) ways and a second event can occur in \(n\) ways,
then both events can occur in \(m \times n\) ways
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\(3 \text{ types of }roti \ \times 2 \text{ types of gravy } = 6 \text{ ways to choose a breakfast set}\) |
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- This rule can also be applied to more than two events
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Example |
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(a) |
Determine the number of ways to toss a dice and a piece of coin simultaneously.
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(b) |
Find the number of ways a person can guess a \(4\)-digit code to access a cell phone if the digits can be repeated.
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Penyelesaian: |
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(a) |
A dice has \(6\) surfaces and a piece of coin has \(2\) surfaces.
Hence, the number of ways to toss both objects simultaneously is
\(6 × 2 = 12\).
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(b) |
The number of ways a person can guess the \(4\)-digit code to access a cell phone is
\(10 \times 10 \times 10 \times 10 = 10 \ 000\) because there are \(10\) digits of number.
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Permutation |
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The number of ways to arrange these letters.
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- Determining the number of permutation for \(n\) different objects:
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\(n! = \ _{n}P_{r} = n \times (n-1) \times (n-2) \times \ ... \ \times 3 \times 2 \times 1\)
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- The permutation of \(n\) objects in a circle is:
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\(P = \dfrac{n!}{n} = \dfrac{n(n-1)!}{n} = (n-1)!\) |
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Example |
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(a) |
Find the value of \(\dfrac{11!}{9!}\) without using a calculator. |
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(b) |
Find the number of ways to arrange all the letters from the word BIJAK when repetition of letters is not allowed. |
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(c) |
Determine the number of ways to arrange six pupils to sit at a round table. |
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Solution: |
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(a) |
\(\begin{aligned} \dfrac{11!}{9!} &= \dfrac{11 \times 10 \times 9!}{9!}\\\\ &=11\times 10\\\\ &=110 \end{aligned}\) |
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(b) |
Given the number of letters, \(n=5\).
Thus, the number of ways to arrange all the letters is \(5! = 120\).
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(c) |
Given the number of pupils, \(n=6\).
Thus, the number of ways to arrange the six pupils is
\((6 – 1)! = 120\).
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- Determining the number of permutation of \(n\) different objects, taking \(r\) objects each time:
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\( _{n}P_{r} = \dfrac{n!}{(n-r)!}\), where \(r \leqslant n\) |
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- The number of permutations for \(n\) different objects taking \(r\) objects each time and arranged in a circle is given by:
Remark |
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A permutation of an object in a circle where clockwise and anticlockwise arrangements are the same, then the number of ways is
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\(\dfrac{_{n}P_{r}}{2r}\) |
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Example |
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(a) |
Eight committee members from a society are nominated to contest for the posts of President, Vice President and Secretary.
How many ways can this three posts be filled?
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(b) |
Nadia bought \(12\) beads of different colours from Handicraft Market in Kota Kinabalu and she intends to make a bracelet.
Nadia realises that the bracelet requires only \(8\) beads.
How many ways are there to make the bracelet?
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Solution: |
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(a) |
Three out of the eight committee members will fill up the three posts.
Hence, \(_{8}P_{3} = \dfrac{8!}{(8-3)!} = 336 \)
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(b) |
Given the number of beads is 12 and 8 beads are to be arranged to form a bracelet.
It is found that clockwise and anticlockwise arrangements are identical.
Hence, the number of permutations is
\(\dfrac{_{12}P_{8}}{2(8)} = 1 \ 247 \ 400\)
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- Determining the number of permutations for \(n\) objects involving identical objects:
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\(P = \dfrac{n!}{a!b!c! \ ...}\), where |
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\(a, b \text{ and } c, \ ... \) are the number of identical objects for each type |
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Example |
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Calculate the number of ways to arrange the letters from the word SIMBIOSIS
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Solution: |
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Given \(n=9\).
The identical objects for letters S and I are the same, which is \(3\).
Hence, the number of ways to arrange the letters from the word SIMBIOSIS is
\(\dfrac{9!}{3!3!}= 10\ 080\)
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