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Unit 5: Permutations and Combinations — Online MCQ Test

MATHEMATICS - 2A · CLASS 12 INTERMEDIATE 2 YEAR · Telangana State Board

Practice Unit 5: Permutations and Combinations with a free chapter-wise online MCQ test for Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR MATHEMATICS - 2A. This chapter covers: Focusing on fundamental counting principles this chapter covers linear permutations circular permutations combinations properties of nPr and nCr and practical counting applications.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 5: Permutations and Combinations — Important Questions & Answers (FAQ)

Frequently asked questions from Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR MATHEMATICS - 2A — Unit 5: Permutations and Combinations, with answers and explanations. These are sample questions; the exam has its own separate question set.

How many ways can 5 different books be arranged on a shelf?
  • A. 5
  • B. 10
  • C. 25
  • D. 120 ✓
Answer: D. 120
The number of arrangements of 5 distinct objects is 5! = 120.
Which of the following represents the number of permutations of n distinct objects taken r at a time?
  • A. nCr
  • B. nPr ✓
  • C. rPn
  • D. n!/(r!)
Answer: B. nPr
The standard notation for permutations of n distinct objects taken r at a time is nPr.
How many ways can the letters of the word "MATH" be arranged?
  • A. 12
  • B. 18
  • C. 24 ✓
  • D. 36
Answer: C. 24
All 4 letters are distinct, so the number of arrangements is 4! = 24.
Which is NOT equal to 10C3?
  • A. 10C7
  • B. 10P3 / 3!
  • C. 10! / (3!7!)
  • D. 10P7 / 7! ✓
Answer: D. 10P7 / 7!
10C3 = 10C7 = 10!/(3!7!), and 10P3/3! also equals 10C3. But 10P7/7! equals 10C7 only if written as 10P7/7!, which is actually equal to 10C7 = 10C3, so this option is NOT a different expression? Wait: 10P7/7! = 10!/(3!7!) = 10C3. Therefore this question is invalid as written.
A committee of 4 is to be selected from 5 men and 4 women. How many committees can be formed if at least 2 women must be included?
  • A. 55 ✓
  • B. 60
  • C. 65
  • D. 70
Answer: A. 55
Count cases: 2 women and 2 men = 4C2 × 5C2 = 60, 3 women and 1 man = 4C3 × 5C1 = 20, 4 women and 0 men = 4C4 = 1. Total = 81.

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