Ex 7.3, 7 (i) - Chapter 7 Class 11 Permutations and Combinations (Term 2)

Transcript

Ex 7.3, 7 Find r if 5Pr = 2 6Pr – 1 Calculating 5Pr & 6Pr – 1 Given, 5Pr = 26Pr-1 5Pr = 5!/(5 − 𝑟)! 6Pr – 1 = 6!/(6 −(𝑟−1))! = 6!/(6 − 𝑟 + 1)! = 6!/(7 − 𝑟)! nPr = 𝑛!/(𝑛 − 𝑟)! 5!/(5 − 𝑟)! = 2 × 6!/(7 − 𝑟)! ((7 − 𝑟)! )/(5 − 𝑟)! = 2 × 6!/5! ((7 − 𝑟)(6 − 𝑟)(5 − 𝑟)! )/(5 − 𝑟)! = 2 × 6!/5! (7 – r)(6 – r) = 2 × 6!/5! (7 – r)(6 – r) = 2 × (6 × 5!)/5! (7 – r)(6 – r) = 2 × 6 (7 – r)(6 – r) = 12 7(6 – r) – r(6 – r) = 12 42 – 7r – 6r + r2 = 12 r2 – 13r + 42 – 12 = 0 r2 – 13r + 30 = 0 r2 – 3r – 10r + 30 = 0 r (r – 3) – 10 (r – 3) = 0 (r – 3) (r – 10) = 0 Hence, r = 3, 10 But, r < n So, r < 5 and r < 6 ∴ r = 10 is not possible So, r = 3 is the answer