10 Permutation and Combination
10 Permutation and Combination
Review Exercise 10 (p. 10.3)
1.
From the tree diagram, there are 8 possible outcomes.
2.
From the tree diagram, there are 6 possible choices.
3. / Multiple-choice questionA / B / C / D
True-false question / T / T, A / T, B / T, C / T, D
F / F, A / F, B / F, C / F, D
From the table above, there are 8 possible outcomes.
4. / 2nd digit1 / 2 / 3
1st digit / 1 / 11 / 12 / 13
2 / 21 / 22 / 23
3 / 31 / 32 / 33
From the table above, there are 9 possible outcomes.
Activity
Activity 10.1 (p. 10.17)
1. (b)
(c)
2. (a)
(b)
(c)
(d)
3. From the results of questions 1 and 2, we can see that the number of permutations of n distinct objects taken r at a time is a product of r consecutive terms.
Activity 10.2 (p. 10.29)
1. (a) The number of permutations of 4 coloured cups taken 2 at a time is 12 .
(b) / Permutation / / CombinationRB / BR / / RB
RG / GR / / RG
RO / OR / / RO
BG / GB / / BG
BO / OB / / BO
GO / OG / GO
From the table above, we can see that each combination of 2 colours can be permuted in
2! = 2 ways.
Therefore, the number of combinations of 2 coloured cups selected from 4 is 6 .
2. (a) The number of permutations of 4 coloured cups taken 3 at a time is 24 .
(b) / Permutation / / CombinationRBG / RGB / BRG / BGR / GRB / GBR / RBG
RBO / ROB / BRO / BOR / ORB / OBR / RBO
RGO / ROG / GRO / GOR / ORG / OGR / RGO
BGO / BOG / GBO / GOB / OBG / OGB / BGO
From the table above, we can see that each combination
of 3 colours can be permuted in 3! = 6 ways.
Therefore, the number of combinations of 3 coloured
cups selected from 4 is 4 .
3. (a) (b) (c)
Maths Dialogue
Maths Dialogue (p. 10.25)
Ken forgot to remove the arrangements that ‘exactly 2 birds are placed next to each other’ from 7!. Angel forgot to consider that the birds could be placed in the front or at the back of the mammals. The correct number of arrangements is.
Classwork
Classwork (p. 10.14)
1. (a)
(b)
(c)
(d)
2. (a) û (b) û (c) û (d) û
Classwork (p. 10.17)
(a)
(b)
(c)
(d)
(e)
Classwork (p. 10.36)
(a) combination (b) permutation
(c) combination (d) permutation
(e) permutation
Quick Practice
Quick Practice 10.1 (p. 10.5)
By the addition rule of counting,
the number of choices
Quick Practice 10.2 (p. 10.6)
By the addition rule of counting,
the total number of students
Quick Practice 10.3 (p. 10.7)
By the multiplication rule of counting,
the number of different choices
Quick Practice 10.4 (p. 10.8)
(a) By the multiplication rule of counting,
the number of possible PINs
(b) By the multiplication rule of counting,
the number of possible PINs
Quick Practice 10.5 (p. 10.9)
By the multiplication rule of counting,
the number of ways to travel from Quarry Bay to Tuen Mun via
Central
The number of ways to travel directly from Quarry Bay to Tuen Mun
By the addition rule of counting,
the total number of ways to travel from Quarry Bay to Tuen Mun
Quick Practice 10.6 (p. 10.10)
(a) By the addition rule of counting,
the number of choices
(b) By the multiplication rule of counting,
the number of choices
(c) By the multiplication rule of counting,
the number of ways to take a tennis course and a volleyball course
the number of ways to take a volleyball course and a squash course
the number of ways to take a tennis course and a squash course
By the addition rule of counting,
the number of choices
Quick Practice 10.7 (p. 10.15)
(a)
(b)
Quick Practice 10.8 (p. 10.19)
(a)
(b)
Quick Practice 10.9 (p. 10.20)
(a) The number of permutations
(b) The number of permutations
Quick Practice 10.10 (p. 10.21)
(a) The number of 3-digit numbers formed =
(b) The first digit must be 5, 6 or 8.
The number of ways of filling in the first digit is
After choosing 5, 6 or 8 as the first digit, we have
3 numbers left.
The number of ways of filling in the remaining 2 digits
is
By the multiplication rule of counting,
the number of 3-digit numbers formed in (a) are greater than 500
Quick Practice 10.11 (p. 10.22)
(a) The number of ways to choose the 2 letters
The number of ways to choose the 4 digits
By the multiplication rule of counting,
the number of possible passwords =
Quick Practice 10.12 (p. 10.23)
(a) The number of ways of arranging face cards on the right
= 3!
The number of ways of arranging number cards on the left
= 3!
By the multiplication rule of counting,
the number of possible arrangements
(b) We can treat all the face cards as a unit.
The number of ways of arranging the number cards and the unit = (3 + 1)! = 4!
The number of ways of arranging the number cards = 3!
By the multiplication rule of counting,
the number of possible arrangements
Quick Practice 10.13 (p. 10.24)
Arrange the consonants B, C, D and F in a row.
The number of ways to arrange the 4 consonants is 4!.
Since vowels A and E must be separated by consonants, they can be arranged into the 5 positions between or next to the consonants as shown below.
Consonants:Vowels: / / / / /
The number of ways to arrange the vowels
∴ The required number of permutations =
Alternative Solution
The total number of permutations without restrictions
= 6! = 720
The number of permutations with the vowels next to each other = 5! × 2! = 240
∴ The required number of permutations
Quick Practice 10.14 (p. 10.31)
(a)
(b)
Quick Practice 10.15 (p. 10.32)
(a) The required number of ways =
(b) The required number of ways =
Quick Practice 10.16 (p. 10.34)
(a) The number of ways of selecting 3 girls and 2 boys
(b) Since 2 particular girls must be included and 4 particular boys must be excluded, there are (12 – 2) = 10 girls and (10 – 4) = 6 boys remained for selection.
The required number of ways
Quick Practice 10.17 (p. 10.35)
(a) The total number of mathematicians and physicists
The required number of committees formed
(b) Exactly 4 mathematicians must be included.
The required number of committees formed
(c) The committee can have 1 mathematician and 7 physicists, or 2 mathematicians and 6 physicists, or 3 mathematicians and 5 physicists.
The number of committees with exactly 1 mathematician
The number of committees with exactly 2 mathematicians
The number of committees with exactly 3 mathematicians
By the addition rule of counting,
the required number of committees formed
(d) The number of committees with more than 3 physicists
The required number of committees formed
Quick Practice 10.18 (p. 10.37)
Number of ways of selecting 9 boys from 15 boys
Number of ways of selecting a prince and a princess role from the 9 selected children
∴ The required number of ways
Quick Practice 10.19 (p. 10.38)
(a) The number of ways to arrange 4 girls in the front row = 4!
The number of ways to arrange 6 boys in the back row = 6!
∴ The number of possible arrangements
(b) (i) The number of possible arrangements
(ii) Consider the girls stand in the first row.
The number of ways of selecting a boy to stand in the first row
The number of ways to arrange 4 girls and 1 boy in the first row
The number of ways to arrange the remaining 5 boys in the second row
The number of ways to arrange the boys and girls if the 4 girls stand in the first row
The situation is similar if the girls stand in the back row.
The number of ways of arranging the boys and girls if the 4 girls stand in the back row
∴ The total number of arrangements
Further Practice
Further Practice (p. 10.6)
1. By the addition rule of counting,
the number of possible ways
2. Number of ways of selecting an even number = 10
Number of ways of selecting a multiple of 3
(i.e. 3, 6, 9, 12, 15, 18) = 6
Number of ways of selecting an even multiple of 3
(i.e. 6, 12, 18) = 3
By the addition rule of counting,
the required number of ways
3. (a) Number of ways of choosing an adult = 4
Number of ways of choosing a female = 1 + 1 + 2 = 4
Number of ways of choosing a female adult = 2
∴ The required number of ways
(b) Number of ways of choosing a child = 3 + 2 = 5
Number of ways of choosing a male = 1 + 1 + 3 = 5
Number of ways of choosing a male child = 3
∴ The required number of ways
Further Practice (p. 10.10)
1. By the multiplication rule of counting,
the number of choices with 2 models having both 3G and 4G versions = 2 × 4 × 2 = 16
The number of choices with the model having a 3G version only = 4
By the addition rule of counting,
the required number of choices
2. (a) By the multiplication rule of counting,
the required number of ways
(b) By the addition rule of counting,
the required number of ways
(c) By the multiplication rule of counting,
the required number of ways
3. (a) The number of choices of □ (i.e. 0, 5) = 2
The number of choices of « = 10
By the multiplication rule of counting,
the number of ways of forming the 3-digit number
(b) The number of choices of □ (i.e. 0, 5) = 2
The number of choices of «(i.e. except 4 and □)
= 10 – 1 – 1 = 8
By the multiplication rule of counting,
the number of ways of forming the 3-digit number
Further Practice (p. 10.26)
1. (a) The number of 4-letter strings formed =
(b) The string must begin with A and end with O, or vice versa.
The number of arrangements for the first and the last letters is 2!.
Then 3 letters remain for the 2nd and 3rd letters.
The number of ways of arranging the 2nd and the 3rd letters is.
∴ The required number of the 4-letter strings
2. (a) (i) The total number of singers = 4 + 5 = 9
The required number of performing sequences
(ii) The number of ways to select the first and the last male singers is.
The number of ways to arrange the remaining
7 singers = 7!
The required number of performing sequences
(b) The total number of singers = 3 + 3 = 6
If the first singer is a male, the number of ways to arrange the male singers is 3!.
The number of ways to arrange the female singers is also.
By the multiplication rule of counting,
the number of ways to arrange the singers if the first singer is a male
The situation is similar if the first singer is a female.
The number of ways to arrange the singers if the first singer is a female
Thus, the required number of ways
3. (a) The number of ways to arrange phones D and E = 2!
The number of ways to arrange phones A, B, C, F, G and H = 6!
By the multiplication rule of counting,
the required number of ways
(b) We can treat phones of the same brand as one unit.
The number of ways to arrange the 3 units = 3!
The number of ways to arrange the brand I phones
= 3!
The number of ways to arrange the brand II phones
= 2!
The number of ways to arrange the brand III phones
= 3!
By the multiplication rule of counting,
the required number of ways
(c) Phone D can be inserted into the 6 positions between or next to the 6 phones as shown such that phone E can then be placed and separated by exactly 4 phones from phone D.
/ / / / / For each position of phone D, there is only one corresponding position for phone E.
The number of ways to arrange phones D and E = 6
The number of ways to arrange the other 6 phones = 6!
∴ The required number of ways
Further Practice (p. 10.38)
1. (a) The required number of choices
(b) The required number of choices