Permutation and Combination questions with answers and concepts

Learn permutation and combination with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

135 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Permutation and Combination are fundamental concepts in counting problems. Permutation deals with arrangements where order matters, while Combination deals with selections where order doesn't matter. These concepts are essential for solving problems involving arrangements, selections, and probability calculations.

Key formulas and rules

9 rules

Permutation (arrangements, order matters): $^{n} P_{r} = \frac{n !}{( n - r )!}$

Combination (selections, order does not matter): $^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$

Relationship: $^{n} P_{r} =^{n} C_{r} \times r!$

Symmetry property: $^{n} C_{r} =^{n} C_{n - r}$

Pascal's Identity: $^{n} C_{r} =^{n - 1} C_{r - 1} +^{n - 1} C_{r}$

Total subsets (sum of combinations): $\sum_{r = 0}^{n} (r n) = 2^{n}$

Arrangements with identical items: $\frac{n !}{p ! \times q ! \times r !}$

Circular permutation (distinct items): $(n - 1)!$

Circular permutation (garland/necklace, reversible): $\frac{( n - 1 )!}{2}$

Key concepts

Fundamental Principle of Counting

Multiplication Principle: If one operation can be performed in 'm' ways and another in 'n' ways independently, then both operations together can be performed in m x n ways. Addition Principle: If two operations are mutually exclusive (either one OR the other), the total ways = m + n ways. Always determine whether to multiply (AND) or add (OR) based on the problem structure.

Permutations - Order Matters

Use permutations when the arrangement or order of selection is important. For example: forming numbers, arranging people in a line, assigning distinct positions, creating codes with different arrangements. Key scenarios: linear arrangements (nPr), circular arrangements ((n-1)!), arrangements with repetition $(n^{r})$ , and arrangements with identical items (accounting for duplicates by dividing by factorial of identical counts).

Combinations - Order Doesn't Matter

Use combinations when only the selection matters, not the order. For example: forming committees, selecting team members, choosing items from a menu, drawing cards from a deck. The formula nCr automatically accounts for the fact that different arrangements of the same items should be counted as one. Always verify whether the problem involves selection only (combination) or arrangement (permutation).

Special Cases and Constraints

Conditional arrangements: When certain items must or must not be together, use the gap method or treat restricted items as a single unit. At least/at most problems: Often easier to calculate using complement - subtract unwanted cases from total. Derangements: Ways to arrange n items so none appears in its original position. Distributing identical items into distinct groups: Use 'stars and bars' method.

Circular and Restricted Arrangements

Circular arrangements have (n-1)! ways because rotations of the same arrangement are considered identical. For arrangements where clockwise and anticlockwise are the same (like necklace beads), divide by 2: (n-1)!/2. When arranging with restrictions (certain people not sitting together), first arrange unrestricted items, then place restricted items in the gaps formed.

Grouping and Distribution Problems

Dividing n distinct items into groups of sizes a, b, c... where groups are identical in size: divide by the factorial of identical group counts to avoid overcounting. Distribution into distinct boxes: Multiply combinations for each box. When distributing identical items into distinct boxes with no restrictions: use (n+r-1)C(r-1) where n=items, r=boxes.

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

In how many distinct ways can 5 people be seated around a circular table?

Solution

Step 1: Apply circular permutation formula $(n - 1)!$ .
In a circular arrangement, rotation creates identical relative configurations, so 1 position is fixed.

P_{circ} = (5 - 1)! = 4!

Step 2: Evaluate factorial.

4! = 4 \times 3 \times 2 \times 1 = 24

Answer: 24 ways.

Example 2

A committee of 4 members is to be selected from 8 men and 6 women. In how many ways can this be done if the committee must include at least 2 women?

Solution

Step 1: Partition into mutually exclusive cases with $\geq 2$ women.
- Case 1 (2 women, 2 men): $(2 6) \times (2 8) = 15 \times 28 = 420$
- Case 2 (3 women, 1 man): $(3 6) \times (1 8) = 20 \times 8 = 160$
- Case 3 (4 women, 0 men): $(4 6) \times (0 8) = 15 \times 1 = 15$

Step 2: Sum all valid cases.

Total = 420 + 160 + 15 = 595

Answer: 595 ways.

Example 3

How many distinct permutations can be formed using all the letters in the word 'ARRANGE'?

Solution

Step 1: Count letters and repetitions.
- Total letters $n = 7$
- Letter 'A' repeats 2 times
- Letter 'R' repeats 2 times

Step 2: Apply multi-set permutation formula.

Permutations = \frac{n !}{p ! q !} = \frac{7 !}{2 ! \times 2 !} = \frac{5 , 040}{2 \times 2} = \frac{5 , 040}{4} = 1, 260

Answer: 1,260 words.

Example 4

There are 10 people in a conference room. If every person shakes hands with every other person exactly once, how many total handshakes occur?

Solution

Step 1: A handshake requires selecting a pair of 2 individuals out of 10.

Handshakes = (2 10) = \frac{10 \times 9}{2 \times 1} = 45

Answer: 45 handshakes.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

How many 4-digit numbers can be formed using digits 0, 1, 2, 3, 4, 5 without repetition?

Answer: B. 300

Explanation

Step 1: First digit cannot be 0 (otherwise not 4-digit).
Step 2: First digit: 5 choices (1,2,3,4,5).
Step 3: Remaining 3 digits from remaining 5: $5 P 3 = 60$ .
Step 1: First digit cannot be 0, so there are 5 choices
Step 2: Fill the remaining 3 places from the remaining 5 digits: 5 x 4 x 3
Step 3: Total = 5 x 5 x 4 x $3 = 300$
Answer: 300

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always first determine if the problem is about permutation (order matters) or combination (order doesn't matter).
For circular arrangements, remember the formula is (n-1)!, not n!. When clockwise/anticlockwise are same, use (n-1)!/2.
Use the complement method for 'at least' or 'at most' problems - subtract unwanted cases from total cases.
When dealing with identical items, divide by the factorial of the count of each identical item type to account for indistinguishable arrangements.
Treat items that must stay together as a single unit, arrange the units, then arrange within the unit.
For 'not together' problems, first arrange the other items, then place the restricted items in the gaps formed.
Remember: nCr = nC(n-r). Use this to simplify calculations when r is large (e.g., 20C18 = 20C2 = 190).

Ready to test your understanding?

Work through 135 questions with explanations after each answer.

Start practice

Next topic: Probability

Permutation and Combination questions with answers and concepts

135 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Key formulas and rules

9 rules

Permutation (arrangements, order matters): $^{n} P_{r} = \frac{n !}{( n - r )!}$

Combination (selections, order does not matter): $^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$

Relationship: $^{n} P_{r} =^{n} C_{r} \times r!$

Symmetry property: $^{n} C_{r} =^{n} C_{n - r}$

Pascal's Identity: $^{n} C_{r} =^{n - 1} C_{r - 1} +^{n - 1} C_{r}$

Total subsets (sum of combinations): $\sum_{r = 0}^{n} (r n) = 2^{n}$

Arrangements with identical items: $\frac{n !}{p ! \times q ! \times r !}$

Circular permutation (distinct items): $(n - 1)!$

Circular permutation (garland/necklace, reversible): $\frac{( n - 1 )!}{2}$

Key concepts

Fundamental Principle of Counting

Permutations - Order Matters

Combinations - Order Doesn't Matter

Special Cases and Constraints

Circular and Restricted Arrangements

Grouping and Distribution Problems

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

In how many distinct ways can 5 people be seated around a circular table?

Solution

Step 1: Apply circular permutation formula $(n - 1)!$ .
In a circular arrangement, rotation creates identical relative configurations, so 1 position is fixed.

P_{circ} = (5 - 1)! = 4!

Step 2: Evaluate factorial.

4! = 4 \times 3 \times 2 \times 1 = 24

Answer: 24 ways.

Example 2

A committee of 4 members is to be selected from 8 men and 6 women. In how many ways can this be done if the committee must include at least 2 women?

Solution

Step 2: Sum all valid cases.

Total = 420 + 160 + 15 = 595

Answer: 595 ways.

Example 3

How many distinct permutations can be formed using all the letters in the word 'ARRANGE'?

Solution

Step 1: Count letters and repetitions.
- Total letters $n = 7$
- Letter 'A' repeats 2 times
- Letter 'R' repeats 2 times

Step 2: Apply multi-set permutation formula.

Permutations = \frac{n !}{p ! q !} = \frac{7 !}{2 ! \times 2 !} = \frac{5 , 040}{2 \times 2} = \frac{5 , 040}{4} = 1, 260

Answer: 1,260 words.

Example 4

There are 10 people in a conference room. If every person shakes hands with every other person exactly once, how many total handshakes occur?

Solution

Step 1: A handshake requires selecting a pair of 2 individuals out of 10.

Handshakes = (2 10) = \frac{10 \times 9}{2 \times 1} = 45

Answer: 45 handshakes.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

How many 4-digit numbers can be formed using digits 0, 1, 2, 3, 4, 5 without repetition?

Answer: B. 300

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always first determine if the problem is about permutation (order matters) or combination (order doesn't matter).
For circular arrangements, remember the formula is (n-1)!, not n!. When clockwise/anticlockwise are same, use (n-1)!/2.
Use the complement method for 'at least' or 'at most' problems - subtract unwanted cases from total cases.
When dealing with identical items, divide by the factorial of the count of each identical item type to account for indistinguishable arrangements.
Treat items that must stay together as a single unit, arrange the units, then arrange within the unit.
For 'not together' problems, first arrange the other items, then place the restricted items in the gaps formed.
Remember: nCr = nC(n-r). Use this to simplify calculations when r is large (e.g., 20C18 = 20C2 = 190).

Ready to test your understanding?

Work through 135 questions with explanations after each answer.

Start practice

Next topic: Probability

Key formulas and rules

Key concepts

Fundamental Principle of Counting

Permutations - Order Matters

Combinations - Order Doesn't Matter

Special Cases and Constraints

Circular and Restricted Arrangements

Grouping and Distribution Problems

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Fundamental Principle of Counting

Permutations - Order Matters

Combinations - Order Doesn't Matter

Special Cases and Constraints

Circular and Restricted Arrangements

Grouping and Distribution Problems

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.