Civil Service Exam (Professional) Numerical Ability — Permutation & CombinationExam Answer Templates

Examiner tip

Even for 1-mark questions, show the formula to demonstrate understanding

Model answer

5P3 = 5!/(5-3)! = 5!/2! = 120/2 = 60

Answer structure

• Line 1: Apply the permutation formula and calculate [1 mark]

Scoring breakdown

Common mark deductions

• Not showing the formula
• Arithmetic errors
• Wrong final answer

Key phrases to include

• 5P3
• permutation formula
• 60

Examiner tip

Always mention the repetition first to show you understand the constraint

Model answer

The word MANILA has 6 letters with A repeated twice.
Number of arrangements = 6!/2! = 720/2 = 360 ways.

Answer structure

• Line 1: Identify total letters and repetitions [1 mark]
• Line 2: Apply formula for permutations with repetition [1 mark]

Scoring breakdown

Common mark deductions

• Not identifying the repetition
• Using wrong formula
• Calculation errors

Key phrases to include

• 6 letters
• A repeated twice
• 6!/2!
• 360 ways

Examiner tip

Start by explaining why order doesn't matter to justify using combinations

Model answer

Since order of selection doesn't matter, this is a combination problem.
12C5 = 12!/(5!(12-5)!) = 12!/(5!×7!) = (12×11×10×9×8)/(5×4×3×2×1) = 792 ways.

Answer structure

• Line 1: Identify as combination problem [0.5 marks]
• Line 2: Apply combination formula and calculate [1.5 marks]

Scoring breakdown

Common mark deductions

• Using permutation instead of combination
• Arithmetic errors
• Not stating why it's a combination

Key phrases to include

• combination
• order doesn't matter
• 12C5
• 792 ways

Examiner tip

Break down selection into separate groups and multiply - this shows clear logical thinking

Model answer

Given: 6 men, 5 women, committee of 4 with exactly 2 men and 2 women.
Number of ways to select 2 men from 6 = 6C2 = 15
Number of ways to select 2 women from 5 = 5C2 = 10
Total ways = 6C2 × 5C2 = 15 × 10 = 150 ways.

Answer structure

• Line 1: State the given information and requirement [0.5 marks]
• Line 2: Calculate ways to select men [1 mark]
• Line 3: Calculate ways to select women [1 mark]
• Line 4: Apply multiplication principle [0.5 marks]

Scoring breakdown

Common mark deductions

• Not separating men and women selections
• Forgetting multiplication principle
• Arithmetic errors

Key phrases to include

• exactly 2 men and 2 women
• 6C2
• 5C2
• multiplication principle
• 150 ways

Examiner tip

List the choices for each position systematically to avoid confusion

Model answer

Given: Digits 1, 2, 3, 4, 5, 6 (6 digits available)
To form 4-digit numbers without repetition:
First digit: 6 choices
Second digit: 5 choices (one used)
Third digit: 4 choices (two used)
Fourth digit: 3 choices (three used)
Total = 6 × 5 × 4 × 3 = 360 four-digit numbers.

Answer structure

• Line 1: Identify available digits [0.5 marks]
• Line 2: Count choices for each position [1.5 marks]
• Line 3: Apply multiplication principle [1 mark]

Scoring breakdown

Common mark deductions

• Not accounting for decreasing choices
• Allowing repetition
• Wrong multiplication

Key phrases to include

• without repetition
• 6 choices
• 5 choices
• 4 choices
• 3 choices
• 360

Examiner tip

Count each letter carefully and list all repetitions before applying the formula

Model answer

The word PROBABILITY has 11 letters:
B appears 2 times, I appears 2 times, other letters appear once each.
Number of arrangements = 11!/(2! × 2!) = 39916800/(2 × 2) = 39916800/4 = 9979200 ways.

Answer structure

• Line 1: Count total letters and identify repetitions [1 mark]
• Line 2: Apply formula for permutations with repetition [1 mark]
• Line 3: Calculate final answer [1 mark]

Scoring breakdown

Common mark deductions

• Missing some repetitions
• Wrong formula
• Calculation errors

Key phrases to include

• 11 letters
• B appears 2 times
• I appears 2 times
• 11!/(2! × 2!)
• 9979200

Examiner tip

With repetition allowed, each position has the same number of choices as if it were the first

Model answer

For the security code: 2 letters + 3 digits with repetition allowed
Letters: 26 choices for first position, 26 choices for second position
Digits: 10 choices for each of the three positions
Total codes = 26 × 26 × 10 × 10 × 10 = 676 × 1000 = 676,000 different codes.

Answer structure

• Line 1: Identify the structure (2 letters + 3 digits) [0.5 marks]
• Line 2: Count choices for letters with repetition [1 mark]
• Line 3: Count choices for digits with repetition [1 mark]
• Line 4: Apply multiplication principle [0.5 marks]

Scoring breakdown

Common mark deductions

• Assuming no repetition
• Wrong number of letter/digit choices
• Calculation errors

Key phrases to include

• repetition allowed
• 26 choices
• 10 choices
• 676,000 codes

Examiner tip

Always verify your solution with a specific value to ensure correctness

Model answer

Given: nP3 = 6 × nC3

Step 1: Write the formulas
nP3 = n!/(n-3)!
nC3 = n!/[3!(n-3)!]

Step 2: Substitute into the equation
n!/(n-3)! = 6 × n!/[3!(n-3)!]

Step 3: Simplify
n!/(n-3)! = 6 × n!/[6(n-3)!]
n!/(n-3)! = n!/(n-3)!

This gives us: 1 = 1, which means the equation is satisfied for any valid value of n ≥ 3.

Step 4: Verify with a specific value
Let n = 4: 4P3 = 24, 4C3 = 4, and 6 × 4 = 24 ✓

Therefore, n can be any integer ≥ 3.

Answer structure

• Line 1: State the given equation [0.5 marks]
• Lines 2-4: Write both formulas correctly [1 mark]
• Lines 5-7: Substitute formulas into equation [1 mark]
• Lines 8-10: Simplify the equation [1.5 marks]
• Lines 11-12: Verify with example [1 mark]

Scoring breakdown

Common mark deductions

• Wrong formulas
• Algebraic errors
• No verification
• Incomplete solution

Key phrases to include

• nP3 = n!/(n-3)!
• nC3 = n!/[3!(n-3)!]
• simplify
• n ≥ 3
• verify

Examiner tip

List all cases systematically before calculating to avoid missing any possibilities

Model answer

Given: 8 boys, 6 girls, select 11 members with at least 5 boys

Since we need at least 5 boys, the possible combinations are:

• 5 boys and 6 girls
• 6 boys and 5 girls
• 7 boys and 4 girls
• 8 boys and 3 girls

Case 1: 5 boys, 6 girls
Ways = 8C5 × 6C6 = 56 × 1 = 56

Case 2: 6 boys, 5 girls
Ways = 8C6 × 6C5 = 28 × 6 = 168

Case 3: 7 boys, 4 girls
Ways = 8C7 × 6C4 = 8 × 15 = 120

Case 4: 8 boys, 3 girls
Ways = 8C8 × 6C3 = 1 × 20 = 20

Total ways = 56 + 168 + 120 + 20 = 364 ways.

Answer structure

• Line 1: State given information [0.5 marks]
• Lines 2-6: List all possible cases [1 mark]
• Lines 7-14: Calculate each case [2.5 marks]
• Line 15: Sum all cases for final answer [1 mark]

Scoring breakdown

Common mark deductions

• Missing some cases
• Wrong combinations
• Addition errors
• Not showing all work

Key phrases to include

• at least 5 boys
• possible combinations
• 8C5
• 6C6
• 364 ways

Examiner tip

This is a definition that must be memorized - 0! = 1 always

Model answer

0! = 1 (by definition)

Answer structure

• Line 1: State that 0! equals 1 by definition [1 mark]

Scoring breakdown

Common mark deductions

• Saying 0! = 0
• No answer
• Confusion with undefined

Key phrases to include

• 0! = 1
• by definition

Examiner tip

The unit method is key for 'together' problems - always multiply by internal arrangements

Model answer

When 2 specific people must sit together, treat them as one unit.
This gives us 5 units to arrange: 5! = 120 ways
Within their unit, the 2 people can be arranged in 2! = 2 ways
Total arrangements = 5! × 2! = 120 × 2 = 240 ways.

Answer structure

• Line 1: Treat the 2 people as one unit [1 mark]
• Line 2: Arrange the 5 units [1 mark]
• Line 3: Account for internal arrangement and multiply [1 mark]

Scoring breakdown

Common mark deductions

• Not treating as a unit
• Forgetting internal arrangement
• Wrong calculation

Key phrases to include

• treat as one unit
• 5! ways
• 2! ways
• 240 ways

Examiner tip

Remember that n must be a positive integer in combination problems

Model answer

Given: nC2 = 28
Using the formula: nC2 = n!/[2!(n-2)!] = n(n-1)/2
So: n(n-1)/2 = 28
n(n-1) = 56
n² - n - 56 = 0
(n-8)(n+7) = 0
Since n must be positive: n = 8

Answer structure

• Line 1: Set up the equation using combination formula [1 mark]
• Lines 2-6: Solve the quadratic equation [1 mark]

Scoring breakdown

Common mark deductions

• Wrong formula
• Algebraic errors
• Taking negative value

Key phrases to include

• nC2 = n(n-1)/2
• n(n-1) = 56
• n = 8

Examiner tip

Carefully count each letter - making a frequency table can help avoid errors

Model answer

The word STATISTICS has 10 letters:
S appears 3 times, T appears 3 times, I appears 2 times, A appears 1 time, C appears 1 time
Number of arrangements = 10!/(3! × 3! × 2! × 1! × 1!) = 3628800/(6 × 6 × 2 × 1 × 1) = 3628800/72 = 50400 ways.

Answer structure

• Line 1: Count total letters and repetitions [1 mark]
• Line 2: Apply permutation formula with repetition [1 mark]
• Line 3: Calculate final answer [1 mark]

Scoring breakdown

Common mark deductions

• Miscounting letters
• Wrong formula
• Arithmetic errors

Key phrases to include

• 10 letters
• S appears 3 times
• T appears 3 times
• I appears 2 times
• 50400

Examiner tip

When colors don't matter in the question, treat all balls as identical for selection purposes

Model answer

Total balls = 5 red + 4 blue = 9 balls
Since order of selection doesn't matter, we use combinations.
Number of ways to select 3 balls from 9 = 9C3 = 9!/(3!6!) = (9×8×7)/(3×2×1) = 504/6 = 84 ways.

Answer structure

• Line 1: Find total number of balls [0.5 marks]
• Line 2: Apply combination formula [1.5 marks]

Scoring breakdown

Common mark deductions

• Using permutation
• Wrong total count
• Calculation errors

Key phrases to include

• 9 balls total
• combinations
• 9C3
• 84 ways

Examiner tip

The relationship nPr = r! × nCr is crucial for solving such problems efficiently

Model answer

Given: nPr = 720 and nCr = 120

Step 1: Use the relationship nPr = r! × nCr
720 = r! × 120
r! = 720/120 = 6
Therefore: r = 3 (since 3! = 6)

Step 2: Find n using nCr = 120
nC3 = 120
n!/[3!(n-3)!] = 120
n(n-1)(n-2)/6 = 120
n(n-1)(n-2) = 720

Step 3: Solve by testing values
For n = 10: 10 × 9 × 8 = 720 ✓

Step 4: Verify
10P3 = 10 × 9 × 8 = 720 ✓
10C3 = 720/6 = 120 ✓

Therefore: n = 10, r = 3

Answer structure

• Step 1: Use nPr = r! × nCr to find r [1.5 marks]
• Step 2: Set up equation for n using nCr [1.5 marks]
• Step 3: Solve for n [1.5 marks]
• Step 4: Verify the solution [0.5 marks]

Scoring breakdown

Common mark deductions

• Not using the relationship
• Algebraic errors
• No verification
• Wrong final values

Key phrases to include

• nPr = r! × nCr
• r! = 6
• r = 3
• n = 10
• verify

Do

• State the formula briefly
• Show one-line calculation
• Give clear final answer

Don't

• Over-explain basic concepts
• Show unnecessary steps
• Leave answer unclear

Do

• Identify the type of problem first
• Show formula substitution
• Calculate step-by-step

Don't

• Skip the identification step
• Make calculation errors
• Rush through working

Do

• List given information
• Show each calculation step
• Use proper mathematical notation
• State units in final answer

Don't

• Jump to conclusion
• Skip intermediate steps
• Make careless errors

Do

• Break into numbered steps
• Show all algebraic working
• Verify answer with substitution
• State conclusion clearly

Don't

• Rush any step
• Skip verification
• Leave working unclear
• Make multiple calculation errors