Check sibling questions

Example 17 - If nC9  = nC8, find nC17 - Chapter 7 - Examples

Example 17 - Chapter 7 Class 11 Permutations and Combinations - Part 2
Example 17 - Chapter 7 Class 11 Permutations and Combinations - Part 3
Example 17 - Chapter 7 Class 11 Permutations and Combinations - Part 4

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Example 17 (Method 1) If nC9 = nC8, find nC17 . nC9 = nC8 𝑛!/9!(𝑛 βˆ’ 9)! = 𝑛!/(𝑛 βˆ’ 8)!8! 𝑛!(𝑛 βˆ’ 8)!/(𝑛 βˆ’ 9)!𝑛! = (9! )/8! (𝑛 βˆ’ 8)!/((𝑛 βˆ’ 9)! ) = (9! )/8! ((𝑛 βˆ’ 8)(𝑛 βˆ’ 8 βˆ’ 1)!)/((𝑛 βˆ’ 9)! ) = (9 Γ— 8! )/8! ((𝑛 βˆ’ 8)(𝑛 βˆ’ 9)!)/((𝑛 βˆ’ 9)! ) = 9 n – 8 = 9 n = 17 nCr = 𝑛!/π‘Ÿ!(𝑛 βˆ’ π‘Ÿ)! Now we have to find nC17 Putting n = 17 = 17C17 = 17!/17!(17 βˆ’ 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 βˆ’ 17)! = 17!/(17! Γ— 0!) = 17!/(17! Γ— 1) = 1/1 = 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.