Permutations and Combinations Solved Examples (Class 11 Mathematics)
Permutations and combinations numericals develop counting principles and probability foundations. These problems explore arrangements and selections in pra
TL;DR: Permutations and combinations numericals develop counting principles and probability foundations. These problems explore arrangements and selections i…
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Permutations and combinations numericals develop counting principles and probability foundations. These problems explore arrangements and selections in pra
Permutations and Combinations — Solved Numerical Examples (Step by Step)
Example 1: How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
Solution: We need to arrange 3 digits from 5 available digits. Using permutation: P(5,3) = 5! / (5-3)! = 5! / 2! = 120 / 2 = 60.
Example 2: In how many ways can a committee of 3 people be chosen from 8 people?
Solution: We need to select 3 from 8 people (order doesn't matter). Using combination: C(8,3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.
Example 3: How many words can be formed using letters of the word MATHEMATICS (all letters used)?
Solution: Word: MATHEMATICS has 11 letters. M appears 2 times, A appears 2 times, T appears 2 times, others appear once. Number of arrangements = 11! / (2! × 2! × 2!) = 39,916,800 / 8 = 4,989,600.
Example 4: In how many ways can 5 books be arranged on a shelf if 2 specific books must be together?
Solution: Treat 2 books that must be together as one unit. So we have 4 units to arrange: 4! = 24. Within the unit, 2 books can arrange in 2! = 2 ways. Total = 24 × 2 = 48.
Example 5: How many subsets does a set with 5 elements have?
Solution: Each element can either be included or not included in a subset. Total subsets = 2^5 = 32.
Example 6: Find the number of ways to distribute 10 identical balls into 3 distinct boxes.
Solution: This is a stars and bars problem. Number of ways = C(10 + 3 - 1, 3 - 1) = C(12, 2) = 12! / (2! × 10!) = (12 × 11) / 2 = 66.
Tips
- Permutation P(n,r) = n! / (n-r)! — used when order matters.
- Combination C(n,r) = n! / (r! × (n-r)!) — used when order doesn't matter.
- Identify if the problem is about arrangement (permutation) or selection (combination).
Frequently Asked Questions
What is the difference between permutation and combination?
Permutation counts arrangements where order matters (e.g., passwords). Combination counts selections where order doesn't matter (e.g., committees).
Why is C(n,0) = 1?
C(n,0) = n! / (0! × n!) = 1. There is exactly one way to choose nothing: by not selecting any element.
More Mathematics Solved Examples
- Real Numbers (HCF LCM Euclid)
- Pair of Linear Equations in Two Variables
- Circles
- Statistics
- Linear Equations in Two Variables
- Heron's Formula
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