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Example 13 - Chapter 6 Class 11 Permutations and Combinations
Last updated at May 29, 2023 by Teachoo
Examples
Example 1
Example 2
Example 3 Important
Example 4 Important
Example 5
Example 6 (i)
Example 6 (ii)
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12 (i) Important
Example 12 (ii)
Example 13 You are here
Example 14 Important
Example 15
Example 16 Important
Example 17
Example 18
Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Important
Example 23 Important
Example 24 Important
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 13 Find r, if 5 4Pr = 6 4Pr-1 Now it is given that 5 4Pr = 6 5Pr-1 5 Γ 4!/(4 β π)! = 6 Γ 5!/(5β (π β 1))! 5 Γ 4!/(4 β π)! = 6 Γ 5!/(5 β π + 1)! 5 Γ 4!/(4 β π)! = 6 Γ 5!/(6 β π)! (6 β π)!/(4 β π)! = (6 Γ 5!)/(5 Γ 4! ) ((6 β π)(5 β π)(4 β π)!)/(4 β π)! = (6 Γ 5!)/(5 Γ 4! ) (6 β r) (5 β r) = (6 Γ 5!)/(5 Γ 4! ) (6 β r) (5 β r) = (6 Γ 5 Γ 4!)/(5 Γ 4! ) (6 β r)(5 β r) = 6 6(5 β r) β r(5 β r) = 6 30 β 6r β 5r + r2 = 6 30 β 11r + r2 = 6 r2 β 11r + 30 = 6 r2 β 11r + 30 β 6 = 0 r2 β 11r + 24 = 0 r β 8r β 3r + 24 = 0 r(r - 8) β 3(r β 8) = 0 (r β 8) (r β 3) = 0 Hence r = 3, 8. But, r < n So, r < 4 and r < 5 β΄ r = 8 is not possible So, r = 3 is the answer
