Permutations and Combinations — Previous Year Questions (Class 11 Mathematics)
Permutations and combinations count arrangements and selections. Master factorials, nPr, nCr, and their applications.
TL;DR: Permutations and combinations count arrangements and selections. Master factorials, nPr, nCr, and their applications.
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Permutations and combinations count arrangements and selections. Master factorials, nPr, nCr, and their applications.
Permutations and Combinations — Previous Year Questions with Solutions
Q (2023, 2 marks): Find the number of ways to arrange the letters of the word STRANGE.
Answer: Word: STRANGE
Number of letters = 7
All letters are distinct (S, T, R, A, N, G, E).
Number of arrangements = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Final Answer: 5040 ways
Q (2022, 2 marks): In how many ways can 5 books be arranged on a shelf?
Answer: Number of books = 5
We need to arrange all 5 books in different positions.
Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120
Final Answer: 120 ways
Q (2023, 2 marks): Find nP₃ if n = 10.
Answer: nPr = n!/(n-r)!
10P₃ = 10!/(10-3)! = 10!/7!
= (10 × 9 × 8 × 7!)/(7!)
= 10 × 9 × 8
= 720
Final Answer: 10P₃ = 720
Q (2021, 2 marks): In how many ways can 3 items be selected from 7 items?
Answer: Number of ways to select 3 items from 7 items = 7C₃
7C₃ = 7!/(3!(7-3)!) = 7!/(3!4!)
= (7 × 6 × 5 × 4!)/(3! × 4!)
= (7 × 6 × 5)/(3 × 2 × 1)
= (7 × 6 × 5)/6
= 7 × 5
= 35
Final Answer: 35 ways
Q (2022, 3 marks): How many 4-digit numbers can be formed using digits 0, 1, 2, 3, 4, 5 without repetition, such that the first digit is not zero?
Answer: We need to form 4-digit numbers from {0, 1, 2, 3, 4, 5} without repetition.
First digit: Cannot be 0, so we have 5 choices {1, 2, 3, 4, 5}
Second digit: Can be any of the remaining 5 digits (including 0 if not used, excluding the first digit)
Third digit: Can be any of the remaining 4 digits
Fourth digit: Can be any of the remaining 3 digits
Number of ways = 5 × 5 × 4 × 3 = 300
Final Answer: 300 numbers
Q (2023, 3 marks): A committee of 3 men and 2 women is to be formed from 5 men and 4 women. In how many ways can this be done?
Answer: Select 3 men from 5 men: 5C₃
Select 2 women from 4 women: 4C₂
Total ways = 5C₃ × 4C₂
5C₃ = 5!/(3!2!) = (5 × 4)/(2 × 1) = 10
4C₂ = 4!/(2!2!) = (4 × 3)/(2 × 1) = 6
Total ways = 10 × 6 = 60
Final Answer: 60 ways
Frequently Asked Questions
What is the difference between permutations and combinations?
Permutations (nPr) count arrangements where order matters. Example: arranging 3 people in a line. Combinations (nCr) count selections where order doesn't matter. Example: selecting 3 people for a committee. Formula: nPr = n!/(n-r)! and nCr = n!/(r!(n-r)!). Always nPr ≥ nCr.
What does 0! equal and why?
0! = 1 by definition. Mathematically, this is necessary for formulas like nCn = n!/(n!0!) = 1 to work correctly. Combinatorially, there is exactly one way to arrange zero objects (the empty arrangement), justifying 0! = 1.
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