Get live Maths 1-on-1 Classs - Class 6 to 12
Example 17 - Chapter 7 Class 11 Permutations and Combinations (Term 2)
Last updated at March 22, 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
Example 14 Important
Example 15
Example 16 Important
Example 17 You are here
Example 18
Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Important
Example 23 Important
Example 24 Important
Chapter 7 Class 11 Permutations and Combinations
Serial order wise
- Ex 7.1
- Ex 7.2
- Ex 7.3
- Ex 7.4
-
Examples
- Miscellaneous
Transcript
Example 17 (Method 1) If nC9 = nC8, find nC17 . nC9 = nC8 π!/9!(π β 9)! = π!/(π β 8)!8! π!(π β 8)!/(π β 9)!π! = (9! )/8! (π β 8)!/((π β 9)! ) = (9! )/8! ((π β 8)(π β 8 β 1)!)/((π β 9)! ) = (9 Γ 8! )/8! ((π β 8)(π β 9)!)/((π β 9)! ) = 9 n β 8 = 9 n = 17 nCr = π!/π!(π β π)! Now we have to find nC17 Putting n = 17 = 17C17 = 17!/17!(17 β 17)! = 17!/(17! Γ 0!) = 17!/(17! Γ 1) = 1/1 = 1 Example, 17 (Alternative Method) If nC9 = nC8, find nC17 . Given nC9 = 4C8 If nCp = nCq then either p = q or p + q = n p = q 9 = 8 Which is not possible p + q = n 9 + 8 = n n = 17 Now we have to find nC17 Putting n = 17 = 17C17 = 17!/17!(17 β 17)! = 17!/(17! Γ 0!) = 17!/(17! Γ 1) = 1/1 = 1
