Misc 7 - Chapter 6 Class 11 Permutations and Combinations

Last updated at April 16, 2024 by Teachoo

Transcript

Misc 7
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Student is required to attempt total 8 questions
& selecting atleast 3 question from part 1 & part 2
Atleast means minimum
So, she can select
Option 1 3 from Part 1 & 5 from Part 2
Option 2 4 from Part 1 & 4 from Part 2
Option 3 5 from Part 1 & 3 from Part 2
We have to calculate all these combinations separately and then add it
Option 1 – 3 questions from Part 1 and 5 Questions from Part 2
Number of ways selecting = 5C3 × 7C5
= 5!/3!(5 − 3)! × 7!/5!(7 − 5)!
= 5!/(3! 2!) × 7!/(5! 2!)
= (5 × 4 × 3!)/(3! 2!) × (7 × 6 × 5!)/(5! 2!)
= 5 × 2 × 7 × 3
= 210
Option 2 – 4 questions from Part 1 and 4 Questions from Part 2
Number of ways selecting = 5C4 × 7C4
= 5!/4!(5 − 4)! × 7!/4!(7 − 4)!
= 5!/(4! 1!) × 7!/(4! 3!)
= (5 × 4!)/4! × (7 × 6 × 5 × 4!)/(4! 3!)
= 5 × 7 × 5
= 175
Option 3 – 5 questions from Part 1 and 3 Questions from Part 2
Number of ways selecting = 5C5 × 7C3
= 5!/5!(5 − 5)! × 7!/3!(7 − 3)!
= 5!/(5! 0!) × 7!/(3! 4!)
= 5!/5! × (7 × 6 × 5 × 4!)/(3! 4!)
= 1 × 7 × 5
= 35
Hence,
Total ways = 210 + 175 + 35
= 420 ways

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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