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

Marks

1

Topic

Basic Permutation Formula

Difficulty

easy

Template Id

T1

Examiner Tip

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

Model Answer

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

Question Type

very_short_answer

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

Marks

2

Topic

Permutations with Repetition

Difficulty

medium

Template Id

T2

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.

Question Type

short_answer

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

Marks

2

Topic

Basic Combinations

Difficulty

medium

Template Id

T3

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.

Question Type

short_answer

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

Marks

3

Topic

Combinations with Restrictions

Difficulty

medium

Template Id

T4

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.

Question Type

short_answer

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

Marks

3

Topic

Permutations in Number Formation

Difficulty

medium

Template Id

T5

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.

Question Type

short_answer

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

Marks

3

Topic

Permutations with Multiple Repetitions

Difficulty

hard

Template Id

T6

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.

Question Type

short_answer

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

Marks

3

Topic

Applications with Repetition

Difficulty

medium

Template Id

T7

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.

Question Type

short_answer

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

Marks

5

Topic

Equations involving Permutations and Combinations

Difficulty

hard

Template Id

T8

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.

Question Type

long_answer

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

Marks

5

Topic

Complex Selection Problems

Difficulty

hard

Template Id

T9

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.

Question Type

long_answer

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

Marks

1

Topic

Factorial Basics

Difficulty

easy

Template Id

T10

Examiner Tip

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

Model Answer

0! = 1 (by definition)

Question Type

very_short_answer

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

Marks

3

Topic

Restricted Arrangements

Difficulty

medium

Template Id

T11

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.

Question Type

short_answer

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

Marks

2

Topic

Finding n in Combinations

Difficulty

medium

Template Id

T12

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

Question Type

short_answer

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

Marks

3

Topic

Complex Permutations with Repetition

Difficulty

hard

Template Id

T13

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.

Question Type

short_answer

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

Marks

2

Topic

Basic Selection Problems

Difficulty

easy

Template Id

T14

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.

Question Type

short_answer

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

Marks

5

Topic

Finding n and r

Difficulty

hard

Template Id

T15

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

Question Type

long_answer

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