Permutations and combinations are methods for counting the different ways to arrange or select objects. The main difference is whether the order of arrangement matters.

Permutation Formulas

Scenario	Formula	Example Use Case
Arranging n distinct objects	n!	Arranging 5 books on a shelf
Arranging r objects out of n	nPr = n! / (n - r)!	Choosing 3 winners from 10 people
With repetition allowed	n^r	Forming a 4-digit PIN from 10 digits
With identical items	n! / (p! × q! × …)	Rearranging letters in "BALLOON"

Combination Formula

Scenario	Formula	Example Use Case
Selecting r from n items	nCr = n! / [r! × (n − r)!]	Selecting 3 people from a group of 10

How to Identify Whether to Use Permutation or Combination

Ask yourself:

Is order important?

If yes, use permutation.

Is it just selection without regard to order?

If yes, use a combination.

Examples:

Forming a PIN/password → Permutation

Selecting committee members → Combination

Awarding positions (gold/silver/bronze) → Permutation

Choosing toppings for a pizza → Combination

Step-by-Step Solving Approach

Understand the situation clearly. Identify whether it's selection or arrangement.

Check for conditions like repetition, identical items, or circular arrangement.

Choose the correct formula.

Permutation: n!, nPr, n^r, n! / (p! × q!...)

Combination: nCr, with or without restrictions.

Plug in the values and simplify.

Interpret your answer in the context of the problem.

Example Explained

Question:
In how many ways can 3 medals (gold, silver, bronze) be given to 6 athletes?

Step 1: Is order important?
Yes. Gold to A and Silver to B is different from Silver to A and Gold to B.

Step 2: Use Permutation Formula
n = 6, r = 3 → Use 6P3

Step 3: Apply the formula
6P3 = 6! / (6 - 3)! = 720 / 6 = 120 ways

10 Solved Examples (Permutations and Combinations Questions)

Example 1: Permutation - Seating People
Q: How many ways can 4 people be seated in a row from a group of 6?
Solution:
Step 1: Use permutation formula → 6P4 = 6! / (6 - 4)!
Step 2: 6! = 720, (6 - 4)! = 2! = 2
Step 3: 6P4 = 720 / 2 = 360 ways

Example 2: Combination - Selecting Students
Q: How many ways can 3 students be selected from 7?
Solution:
Step 1: Use the combination formula → 7C3 = 7! / (3! × 4!)
Step 2: 7! = 5040, 3! = 6, 4! = 24
Step 3: 7C3 = 5040 / (6 × 24) = 5040 / 144 = 35 ways

Example 3: Permutation with Repetition - 3-letter Codes
Q: How many 3-letter codes can be formed from 5 letters if repetition is allowed?
Solution:
Step 1: Repetition allowed → use formula = 5^3
Step 2: 5^3 = 5 × 5 × 5 = 125 codes

Example 4: Combination with Restriction - Selecting Girls
Q: How many ways can 2 girls be selected from 5, if one specific girl must be included?
Solution:
Step 1: Fix 1 girl → 1 way
Step 2: Choose 1 from remaining 4 → 4C1 = 4 ways

Example 5: Permutation with Identical Items - “BALLOON”
Q: How many ways can the letters of “BALLOON” be arranged?
Solution:
Step 1: Total letters = 7
Step 2: Repeated letters: L(2), O(2)
Step 3: Formula = 7! / (1! × 1! × 2! × 2! × 1!)
Step 4: 7! = 5040, Denominator = 4
Result = 5040 / 4 = 1260 arrangements

Example 6: Combination - Forming Teams
Q: From a group of 10 players, how many 5-member teams can be formed?
Solution:
Step 1: Use formula → 10C5 = 10! / (5! × 5!)
Step 2: 10! = 3628800, 5! = 120
Step 3: 10C5 = 3628800 / (120 × 120) = 3628800 / 14400 = 252 teams

Example 7: Permutation - Awarding Medals
Q: In how many ways can 3 medals be awarded to 8 players?
Solution:
Step 1: Use formula → 8P3 = 8! / (8 - 3)! = 8! / 5!
Step 2: 8! = 40320, 5! = 120
Step 3: 40320 / 120 = 336 ways

Example 8: Permutation - 3-digit Numbers
Q: How many 3-digit numbers can be made from 1, 2, 3, 4, and 5 without repetition?
Solution:
Step 1: Use formula → 5P3 = 5! / (5 - 3)! = 5! / 2!
Step 2: 5! = 120, 2! = 2
Step 3: 120 / 2 = 60 numbers

Example 9: Combination - Choosing Subjects
Q: A student has to choose 3 out of 6 optional subjects. In how many ways can this be done?
Solution:
Step 1: Use formula → 6C3 = 6! / (3! × 3!)
Step 2: 6! = 720, 3! = 6
Step 3: 720 / (6 × 6) = 720 / 36 = 20 ways

Example 10: Combination - Forming a Committee
Q: A committee of 2 men and 2 women is to be formed from 4 men and 5 women.
Solution:
Step 1: Choose 2 men from 4 → 4C2 = 6
Step 2: Choose 2 women from 5 → 5C2 = 10
Step 3: Total ways = 6 × 10 = 60 committees

Practice Exercise - Permutations and Combinations Questions

How many ways can 5 books be arranged on a shelf?
In how many ways can 3 people be selected from 8?
From 6 digits, how many 4-digit PINs can be made if digits repeat?
In how many ways can 3 medals be awarded to 6 athletes?
How many 4-letter words can be formed from “MATH”?
From 10 students, how many teams of 4 can be selected?
A password consists of 3 letters and 2 digits. How many unique passwords can be formed (no repetition)?
How many ways can you select 2 girls and 3 boys from 4 girls and 5 boys?
How many arrangements can be made with the letters of the word “GARDEN”?
In how many ways can 3 vowels be selected from 5 vowels?
How many 3-digit even numbers can be made using digits 1 to 6 with no repetition?
From 9 books, how many ways can you pick and arrange 3?
How many ways can a class monitor and assistant be chosen from 20 students?
In how many ways can 5 students be seated around a round table?
How many committees of 3 members can be formed from 7 people?

Conclusion: Permutations and Combinations Questions

Permutations and combinations are useful tools for counting arrangements and selections. The main difference is that permutations consider order, while combinations do not. By learning when and how to use each, and practicing with real-life problems, students can improve their reasoning and problem-solving skills. With the right approach and formulas, these concepts become easy to grasp and valuable in many fields.

Related Links

Permutation and Combination- Ready to master Permutations and Combinations? Practice with real-world problems and boost your math skills today!

Ascending Order: Learn how to arrange numbers in ascending order with easy tips at Orchids The International School.

Arithmetic Progression: Understand arithmetic progression with simple formulas and examples at Orchids The International School

Frequently Asked Questions on Permutations and Combinations Questions

1. What is permutation and combination?

Ans: Permutation: Arrangement of items where order matters.
Formula: nPr = n! / (n - r)!

Combination: Selection of items where order does not matter.
Formula: nCr = n! / [r! × (n - r)!]

2. What are the possible combinations of 1, 2, 3, 4, 5?

Ans: All combinations (without repetition) from the set {1, 2, 3, 4, 5}:

Size 1: {1}, {2}, {3}, {4}, {5}
Size 2: {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}
Size 3: {1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}
Size 4: {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}
Size 5: {1,2,3,4,5}

Total combinations (excluding empty set): 2⁵ - 1 = 31

3. How many combinations of 2 out of 5?

Ans: Using the combination formula:

5C2 = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 10

Total 10 combinations

Combinations:

{1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}

4. What are the permutations of 1, 2, 3, 4?

Ans: Total permutations = 4! = 24

List:
1234, 1243, 1324, 1342, 1423, 1432
2134, 2143, 2314, 2341, 2413, 2431
3124, 3142, 3214, 3241, 3412, 3421
4123, 4132, 4213, 4231, 4312, 4321

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