SPM KSSM Diagnostic Drill
Form 5 Chapter 4: Permutations & Combinations (Pilihatur dan Gabungan)
Total Marks: 40 Marks • Time Allowed: 50 Minutes
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A Bahagian A: SPM Paper 1 Format [16 Marks]
Answer all questions
Soalan 1
[4 Marks]
A four-digit number is to be formed using the digits $\{1, 3, 5, 7, 8, 9\}$ without repetition.
(a) How many 4-digit numbers can be formed in total? [1 mark]
(b) How many of these four-digit numbers are greater than $5\,000$ and are odd numbers? [3 marks]
(a) $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$
[N1] 360
(b) Greater than $5000 \implies$ first digit must be from $\{5, 7, 8, 9\}$. Odd $\implies$ last digit must be from $\{1, 3, 5, 7, 9\}$.
Since $5, 7, 9$ are both $> 5000$ and odd, we partition by the first digit:
• Case 1: First digit is even (8):
First digit: 1 choice (8). Last digit: 5 choices $\{1, 3, 5, 7, 9\}$. Remaining 2 middle digits from 4 remaining digits: $^4P_2 = 12$.
Ways $= 1 \times 5 \times 12 = 60$. [K1]
First digit: 1 choice (8). Last digit: 5 choices $\{1, 3, 5, 7, 9\}$. Remaining 2 middle digits from 4 remaining digits: $^4P_2 = 12$.
Ways $= 1 \times 5 \times 12 = 60$. [K1]
• Case 2: First digit is odd $\{5, 7, 9\}$:
First digit: 3 choices. Last digit: 4 remaining odd choices. Remaining 2 middle digits: $^4P_2 = 12$.
Ways $= 3 \times 4 \times 12 = 144$. [K1]
First digit: 3 choices. Last digit: 4 remaining odd choices. Remaining 2 middle digits: $^4P_2 = 12$.
Ways $= 3 \times 4 \times 12 = 144$. [K1]
Total $= 60 + 144 = 204$. [N1] 204
Soalan 2
[4 Marks]
Consider the 10 letters in the word S T A T I S T I K S.
(a) Find the number of distinct permutations of all 10 letters. [2 marks]
(b) Find the number of arrangements where all 3 letters 'S' are kept together. [2 marks]
Letter frequency: S: 3, T: 3, I: 2, A: 1, K: 1. Total $= 10$ letters.
(a) $\frac{10!}{3! \times 3! \times 2!} = \frac{3\,628\,800}{6 \times 6 \times 2} = \frac{3\,628\,800}{72} = 50\,400$
[K1] $\frac{10!}{3!3!2!}$[N1] $50\,400$
(b) Bundle the 3 'S's into 1 single block: $[SSS]$.
Total items $= 1 \text{ block} + 7 \text{ other letters} = 8 \text{ items}$.
Other letters contain T: 3, I: 2, A: 1, K: 1.
Arrangements $= \frac{8!}{3! \times 2!} = \frac{40\,320}{6 \times 2} = 3\,360$.
(Internal arrangement of identical S's is $\frac{3!}{3!} = 1$).
[K1] $\frac{8!}{3!2!}$Total items $= 1 \text{ block} + 7 \text{ other letters} = 8 \text{ items}$.
Other letters contain T: 3, I: 2, A: 1, K: 1.
Arrangements $= \frac{8!}{3! \times 2!} = \frac{40\,320}{6 \times 2} = 3\,360$.
(Internal arrangement of identical S's is $\frac{3!}{3!} = 1$).
[N1] $3\,360$
Soalan 3
[4 Marks]
A committee of 6 persons comprising 4 men and 2 women are to sit around a circular table.
(a) Find the number of ways to seat them without any restrictions. [1 mark]
(b) Find the number of seating arrangements in which the 2 women must NOT sit next to each other. [3 marks]
(a) Circular permutation of 6 people: $(6 - 1)! = 5! = 120$ ways.
[N1] 120
(b) Slot method:
First, arrange the 4 men around the circular table: $(4 - 1)! = 3! = 6$ ways. [P1]
This creates 4 distinct spaces between the 4 seated men.
Place the 2 women into these 4 spaces: $^4P_2 = 4 \times 3 = 12$ ways. [K1]
Total arrangements $= 6 \times 12 = 72$ ways. [N1] 72
Soalan 4
[4 Marks]
There are 9 distinct coplanar points. 4 of the points lie on a single straight line, and no other 3 points are collinear.
(a) Calculate the number of straight lines that can be formed by joining any two of these points. [2 marks]
(b) Calculate the number of triangles that can be formed using any three of these points as vertices. [2 marks]
Total points $= 9$. 4 collinear points on line $L$. 5 non-collinear points.
(a) Lines formed $= {^9C_2} - {^4C_2} + 1$
$= 36 - 6 + 1 = 31$ straight lines.
[K1] ${^9C_2} - {^4C_2} + 1$$= 36 - 6 + 1 = 31$ straight lines.
[N1] 31
(b) Any 3 collinear points cannot form a triangle:
Triangles formed $= {^9C_3} - {^4C_3}$
$= 84 - 4 = 80$ triangles.
[K1] ${^9C_3} - {^4C_3}$Triangles formed $= {^9C_3} - {^4C_3}$
$= 84 - 4 = 80$ triangles.
[N1] 80
B Bahagian B: SPM Paper 2 Format [24 Marks]
Detailed solutions required
Soalan 5
[10 Marks]
(a) A university debating club has 7 senior members and 5 junior members. A team of 5 members is to be selected for a national championship. Find the number of ways to form the team if:
• (i) the team must have a majority of senior members. [3 marks]
• (ii) the most senior member and the youngest junior member can only be selected if both are included together. [3 marks]
• (i) the team must have a majority of senior members. [3 marks]
• (ii) the most senior member and the youngest junior member can only be selected if both are included together. [3 marks]
(b) In a VIP cinema hall, 4 couples (4 husbands and 4 wives) purchase tickets for an 8-seat row.
• (i) Find the number of ways to seat them if each husband must sit immediately next to his own wife. [2 marks]
• (ii) Find the number of ways to seat them if men and women must sit alternately in the row. [2 marks]
• (i) Find the number of ways to seat them if each husband must sit immediately next to his own wife. [2 marks]
• (ii) Find the number of ways to seat them if men and women must sit alternately in the row. [2 marks]
(a)(i) Majority Seniors (size 5): 3, 4, or 5 seniors:
• 3 Seniors, 2 Juniors: ${^7C_3} \times {^5C_2} = 35 \times 10 = 350$
• 4 Seniors, 1 Junior: ${^7C_4} \times {^5C_1} = 35 \times 5 = 175$
• 5 Seniors, 0 Juniors: ${^7C_5} \times {^5C_0} = 21 \times 1 = 21$
Total $= 350 + 175 + 21 = 546$ ways. [K1] case sum
[N1] 546
• 3 Seniors, 2 Juniors: ${^7C_3} \times {^5C_2} = 35 \times 10 = 350$
• 4 Seniors, 1 Junior: ${^7C_4} \times {^5C_1} = 35 \times 5 = 175$
• 5 Seniors, 0 Juniors: ${^7C_5} \times {^5C_0} = 21 \times 1 = 21$
Total $= 350 + 175 + 21 = 546$ ways. [K1] case sum
[N1] 546
(a)(ii) Both included OR neither included:
Total members $= 12$. The 2 key individuals: $S_1$ and $J_1$. Remaining pool $= 10$.
• Case 1 (Both included): Need 3 more from 10: ${^{10}C_3} = 120$
• Case 2 (Neither included): Need all 5 from 10: ${^{10}C_5} = 252$
Total $= 120 + 252 = 372$ ways. [K1] ${^{10}C_3} + {^{10}C_5}$
[N1] 372
Total members $= 12$. The 2 key individuals: $S_1$ and $J_1$. Remaining pool $= 10$.
• Case 1 (Both included): Need 3 more from 10: ${^{10}C_3} = 120$
• Case 2 (Neither included): Need all 5 from 10: ${^{10}C_5} = 252$
Total $= 120 + 252 = 372$ ways. [K1] ${^{10}C_3} + {^{10}C_5}$
[N1] 372
(b)(i) Couples seated together in row:
Treat each of the 4 couples as 1 block: $4! = 24$ ways to arrange the blocks.
Each couple can swap seats internally: $2! \times 2! \times 2! \times 2! = 2^4 = 16$.
Total ways $= 4! \times 2^4 = 24 \times 16 = 384$ ways. [K1] $4! \times 2^4$
[N1] 384
Treat each of the 4 couples as 1 block: $4! = 24$ ways to arrange the blocks.
Each couple can swap seats internally: $2! \times 2! \times 2! \times 2! = 2^4 = 16$.
Total ways $= 4! \times 2^4 = 24 \times 16 = 384$ ways. [K1] $4! \times 2^4$
[N1] 384
(b)(ii) Men and women alternate:
Pattern 1: $M\,W\,M\,W\,M\,W\,M\,W \implies 4! \times 4! = 24 \times 24 = 576$
Pattern 2: $W\,M\,W\,M\,W\,M\,W\,M \implies 4! \times 4! = 24 \times 24 = 576$
Total ways $= 576 + 576 = 1\,152$ ways. [K1] $2 \times 4! \times 4!$
[N1] 1,152
Pattern 1: $M\,W\,M\,W\,M\,W\,M\,W \implies 4! \times 4! = 24 \times 24 = 576$
Pattern 2: $W\,M\,W\,M\,W\,M\,W\,M \implies 4! \times 4! = 24 \times 24 = 576$
Total ways $= 576 + 576 = 1\,152$ ways. [K1] $2 \times 4! \times 4!$
[N1] 1,152
Soalan 6 • SPM Paper 2 & KBAT
[14 Marks]
International Diplomatic Protocol & Cyber Protection
ASEAN Round Table Seating & AES-Key Token Generation
Part I (Diplomatic Summit Seating Protocol):
8 delegates from 4 sovereign states attend an ASEAN security summit. There are 2 delegates from Malaysia, 2 from Singapore, 2 from Indonesia, and 2 from Thailand. They are to be seated around an equidistant circular conference table.
(a) If delegates from each country must sit next to their fellow country delegate, calculate the number of possible circular seating arrangements. [4 marks]
(b) If only the 2 delegates from Malaysia must sit together, while the 2 delegates from Singapore must NOT sit next to each other, calculate the number of valid seating arrangements. [4 marks]
(a) If delegates from each country must sit next to their fellow country delegate, calculate the number of possible circular seating arrangements. [4 marks]
(b) If only the 2 delegates from Malaysia must sit together, while the 2 delegates from Singapore must NOT sit next to each other, calculate the number of valid seating arrangements. [4 marks]
Part II (Cryptographic Authorization Key):
A cyber-security team designs a 7-character security key for cloud server access.
The pool of available characters consists of:
• 5 distinct special symbols: $\{\#, \$, \%, \&, *\}$
• 6 distinct digits: $\{2, 3, 5, 6, 7, 8\}$
• 4 distinct uppercase letters: $\{P, Q, R, S\}$
(c) A valid token must have the format: [Symbol] [Letter] [Digit] [Digit] [Digit] [Letter] [Symbol], where no character is repeated within the token. Calculate the number of possible tokens that can be generated. [3 marks]
(d) If an additional policy states that the 3 middle digits must form a number strictly greater than $500$, find how many such valid tokens can be formed. [3 marks]
• 5 distinct special symbols: $\{\#, \$, \%, \&, *\}$
• 6 distinct digits: $\{2, 3, 5, 6, 7, 8\}$
• 4 distinct uppercase letters: $\{P, Q, R, S\}$
(c) A valid token must have the format: [Symbol] [Letter] [Digit] [Digit] [Digit] [Letter] [Symbol], where no character is repeated within the token. Calculate the number of possible tokens that can be generated. [3 marks]
(d) If an additional policy states that the 3 middle digits must form a number strictly greater than $500$, find how many such valid tokens can be formed. [3 marks]
(a) Each country's delegates sit together in a circle:
Treat the 4 countries as 4 blocks around a circle: $(4 - 1)! = 3! = 6$ ways. [K1]
Each of the 4 pairs of delegates can swap seats internally: $2! \times 2! \times 2! \times 2! = 16$. [K1]
Total arrangements $= 6 \times 16 = 96$. [K1] $3! \times 2^4$
[N1] 96
Treat the 4 countries as 4 blocks around a circle: $(4 - 1)! = 3! = 6$ ways. [K1]
Each of the 4 pairs of delegates can swap seats internally: $2! \times 2! \times 2! \times 2! = 16$. [K1]
Total arrangements $= 6 \times 16 = 96$. [K1] $3! \times 2^4$
[N1] 96
(b) Malaysia together, Singapore separated:
First, treat Malaysia pair as 1 single unit $[M_1M_2]$ with internal permutation $2! = 2$.
Other unrestricted individuals: Indonesia (2), Thailand (2) $\implies 4$ individuals.
Units to seat in circle first $= 1 \text{ [MY]} + 4 \text{ [ID, TH]} = 5 \text{ units}$.
Circular arrangement of these 5 units $= (5 - 1)! = 4! = 24$ ways. [K1]
This circle of 5 units creates 5 spaces between them.
Insert the 2 Singapore delegates into these 5 spaces so they never touch: $^5P_2 = 5 \times 4 = 20$ ways. [K1]
Multiply by Malaysia internal arrangement: $24 \times 20 \times 2! = 960$ ways. [K1] $4! \times {^5P_2} \times 2!$
[N1] 960
First, treat Malaysia pair as 1 single unit $[M_1M_2]$ with internal permutation $2! = 2$.
Other unrestricted individuals: Indonesia (2), Thailand (2) $\implies 4$ individuals.
Units to seat in circle first $= 1 \text{ [MY]} + 4 \text{ [ID, TH]} = 5 \text{ units}$.
Circular arrangement of these 5 units $= (5 - 1)! = 4! = 24$ ways. [K1]
This circle of 5 units creates 5 spaces between them.
Insert the 2 Singapore delegates into these 5 spaces so they never touch: $^5P_2 = 5 \times 4 = 20$ ways. [K1]
Multiply by Malaysia internal arrangement: $24 \times 20 \times 2! = 960$ ways. [K1] $4! \times {^5P_2} \times 2!$
[N1] 960
(c) Token Format [S] [L] [D] [D] [D] [L] [S]:
• Symbols at positions 1 and 7: choose and arrange 2 from 5: $^5P_2 = 20$. [P1]
• Letters at positions 2 and 6: choose and arrange 2 from 4: $^4P_2 = 12$. [P1]
• Digits at positions 3, 4, 5: choose and arrange 3 from 6: $^6P_3 = 120$.
Total valid tokens $= 20 \times 12 \times 120 = 28\,800$. [K1] $20 \times 12 \times 120$
[N1] 28,800
• Symbols at positions 1 and 7: choose and arrange 2 from 5: $^5P_2 = 20$. [P1]
• Letters at positions 2 and 6: choose and arrange 2 from 4: $^4P_2 = 12$. [P1]
• Digits at positions 3, 4, 5: choose and arrange 3 from 6: $^6P_3 = 120$.
Total valid tokens $= 20 \times 12 \times 120 = 28\,800$. [K1] $20 \times 12 \times 120$
[N1] 28,800
(d) Middle 3 digits strictly $> 500$:
Digits available: $\{2, 3, 5, 6, 7, 8\}$. For a 3-digit number to be $> 500$, the first digit (pos 3) must be in $\{5, 6, 7, 8\}$ (4 choices).
Remaining 2 digits can be any from the remaining 5 digits: $^5P_2 = 5 \times 4 = 20$.
Total valid 3-digit combinations $> 500 = 4 \times 20 = 80$. [K1]
Multiply with symbol and letter arrangements: $20 \times 12 \times 80 = 19\,200$. [K1] $20 \times 12 \times 80$
[N1] 19,200
Digits available: $\{2, 3, 5, 6, 7, 8\}$. For a 3-digit number to be $> 500$, the first digit (pos 3) must be in $\{5, 6, 7, 8\}$ (4 choices).
Remaining 2 digits can be any from the remaining 5 digits: $^5P_2 = 5 \times 4 = 20$.
Total valid 3-digit combinations $> 500 = 4 \times 20 = 80$. [K1]
Multiply with symbol and letter arrangements: $20 \times 12 \times 80 = 19\,200$. [K1] $20 \times 12 \times 80$
[N1] 19,200