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.
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.
Unit 5: Permutations and Combinations — Important Questions & Answers
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.
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.
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.
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.
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.
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.