Example 24 - Chapter 6 Class 11 Permutations and Combinations
Last updated at April 16, 2024 by Teachoo
Examples
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Example 4 Important
Example 5
Example 6 (i)
Example 6 (ii)
Example 7
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Example 11 Important
Example 12 (i) Important
Example 12 (ii)
Example 13
Example 14 Important
Example 15
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Example 17
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Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Important
Example 23 Important
Example 24 Important You are here
Chapter 6 Class 11 Permutations and Combinations
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
- Ex 6.4
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Examples
- Miscellaneous
Transcript
Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together? The seating arrangement would be as follows Hence total number of ways they can sit = 120 × 120 = 14400 ways 5 girls can sit in any of 5 places Number of ways they can sit = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 5 × 4 × 3 × 2 × 1 = 120 3 boys can sit at any of 6 places marked Number of ways they can sit = 6P3 = 6!/(6 − 3)! = 6!/3! = (6 × 5 × 4 × 3!)/3! = 120
