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  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise

Transcript

Example 17 (Method 1) If nC9 = nC8, find nC17 . nC9 = nC8 ๐‘›!/9!(๐‘› โˆ’ 9)! = ๐‘›!/(๐‘› โˆ’ 8)!8! ๐‘›!(๐‘› โˆ’ 8)!/(๐‘› โˆ’ 9)!๐‘›! = (9! )/8! (๐‘› โˆ’ 8)!/((๐‘› โˆ’ 9)! ) = (9! )/8! ((๐‘› โˆ’ 8)(๐‘› โˆ’ 8 โˆ’ 1)!)/((๐‘› โˆ’ 9)! ) = (9 ร— 8! )/8! ((๐‘› โˆ’ 8)(๐‘› โˆ’ 9)!)/((๐‘› โˆ’ 9)! ) = 9 n โ€“ 8 = 9 n = 17 nCr = ๐‘›!/๐‘Ÿ!(๐‘› โˆ’ ๐‘Ÿ)! Now we have to find nC17 Putting n = 17 = 17C17 = 17!/17!(17 โˆ’ 17)! = 17!/(17! ร— 0!) = 17!/(17! ร— 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 โˆ’ 17)! = 17!/(17! ร— 0!) = 17!/(17! ร— 1) = 1/1 = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.