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Page 1 16. Permutations Exercise 16.1 1. Question Compute (i) (ii) (iii). L.C.M. (6!, 7!, 8!) Answer i. Given = 30 × 29 = 870 ii. Given We know that: 11! = 11 × 10 × 9 × … × 1 10! = 10 × 9 × 8 × … × 1 9! = 9 × 8 × 7 × … × 1 Putting these values, we get, = 110 – 10 Hence, iii. We have to find LCM of 6!, 7!, and 8! We can write as, = 8! = 8 × 7 × 6! = 7! = 7 × 6! = 6! = 6! Therefore, L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ] = 8 × 7 × 6! Hence, LCM is 8! Page 2 16. Permutations Exercise 16.1 1. Question Compute (i) (ii) (iii). L.C.M. (6!, 7!, 8!) Answer i. Given = 30 × 29 = 870 ii. Given We know that: 11! = 11 × 10 × 9 × … × 1 10! = 10 × 9 × 8 × … × 1 9! = 9 × 8 × 7 × … × 1 Putting these values, we get, = 110 – 10 Hence, iii. We have to find LCM of 6!, 7!, and 8! We can write as, = 8! = 8 × 7 × 6! = 7! = 7 × 6! = 6! = 6! Therefore, L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ] = 8 × 7 × 6! Hence, LCM is 8! 2. Question Prove that Answer Given: L.H.S Hence, L.H.S = R.H.S 3 A. Question Find x in each of the following : Answer We have We know that 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 Hence, x = 36 3 B. Question Find x in each of the following : Answer Page 3 16. Permutations Exercise 16.1 1. Question Compute (i) (ii) (iii). L.C.M. (6!, 7!, 8!) Answer i. Given = 30 × 29 = 870 ii. Given We know that: 11! = 11 × 10 × 9 × … × 1 10! = 10 × 9 × 8 × … × 1 9! = 9 × 8 × 7 × … × 1 Putting these values, we get, = 110 – 10 Hence, iii. We have to find LCM of 6!, 7!, and 8! We can write as, = 8! = 8 × 7 × 6! = 7! = 7 × 6! = 6! = 6! Therefore, L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ] = 8 × 7 × 6! Hence, LCM is 8! 2. Question Prove that Answer Given: L.H.S Hence, L.H.S = R.H.S 3 A. Question Find x in each of the following : Answer We have We know that 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 Hence, x = 36 3 B. Question Find x in each of the following : Answer We have We can write as 10! = 10 × 9! 9! = 9 × 8! By putting these values Hence, x = 100 3 C. Question Find x in each of the following : Answer We have We can write as. 8! = 8 × 7 × 6! 7! = 7 × 6! Putting the value, we get Hence, x = 64 4 A. Question Page 4 16. Permutations Exercise 16.1 1. Question Compute (i) (ii) (iii). L.C.M. (6!, 7!, 8!) Answer i. Given = 30 × 29 = 870 ii. Given We know that: 11! = 11 × 10 × 9 × … × 1 10! = 10 × 9 × 8 × … × 1 9! = 9 × 8 × 7 × … × 1 Putting these values, we get, = 110 – 10 Hence, iii. We have to find LCM of 6!, 7!, and 8! We can write as, = 8! = 8 × 7 × 6! = 7! = 7 × 6! = 6! = 6! Therefore, L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ] = 8 × 7 × 6! Hence, LCM is 8! 2. Question Prove that Answer Given: L.H.S Hence, L.H.S = R.H.S 3 A. Question Find x in each of the following : Answer We have We know that 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 Hence, x = 36 3 B. Question Find x in each of the following : Answer We have We can write as 10! = 10 × 9! 9! = 9 × 8! By putting these values Hence, x = 100 3 C. Question Find x in each of the following : Answer We have We can write as. 8! = 8 × 7 × 6! 7! = 7 × 6! Putting the value, we get Hence, x = 64 4 A. Question Convert the following products into factorials : 5 · 6 · 7 · 8 · 9 · 10 Answer Given 5 . 6 . 7 . 8 . 9 . 10 Find: Convert into Factorial Now, We can write as = Hence, 4 B. Question Convert the following products into factorials : 3 · 6 · 9 · 12 · 15 · 18 Answer Given 3 . 6 . 9 . 12 . 15 . 18 Find: Convert into Factorial = (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6) = 3 6 (1 × 2 × 3 × 4 × 5 × 6) = 3 6 (6!) 4 C. Question Convert the following products into factorials : (n + 1)(n + 2) (n + 3) …(2n) Answer Given (n + 1)(n + 2) (n + 3) …(2n) Find: Convert into Factorial 4 D. Question Convert the following products into factorials : 1 · 3 · 5 · 7 · 9 …(2n – 1) Answer Given 1 · 3 · 5 · 7 · 9 …(2n – 1) Find: Convert into Factorial Page 5 16. Permutations Exercise 16.1 1. Question Compute (i) (ii) (iii). L.C.M. (6!, 7!, 8!) Answer i. Given = 30 × 29 = 870 ii. Given We know that: 11! = 11 × 10 × 9 × … × 1 10! = 10 × 9 × 8 × … × 1 9! = 9 × 8 × 7 × … × 1 Putting these values, we get, = 110 – 10 Hence, iii. We have to find LCM of 6!, 7!, and 8! We can write as, = 8! = 8 × 7 × 6! = 7! = 7 × 6! = 6! = 6! Therefore, L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ] = 8 × 7 × 6! Hence, LCM is 8! 2. Question Prove that Answer Given: L.H.S Hence, L.H.S = R.H.S 3 A. Question Find x in each of the following : Answer We have We know that 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 Hence, x = 36 3 B. Question Find x in each of the following : Answer We have We can write as 10! = 10 × 9! 9! = 9 × 8! By putting these values Hence, x = 100 3 C. Question Find x in each of the following : Answer We have We can write as. 8! = 8 × 7 × 6! 7! = 7 × 6! Putting the value, we get Hence, x = 64 4 A. Question Convert the following products into factorials : 5 · 6 · 7 · 8 · 9 · 10 Answer Given 5 . 6 . 7 . 8 . 9 . 10 Find: Convert into Factorial Now, We can write as = Hence, 4 B. Question Convert the following products into factorials : 3 · 6 · 9 · 12 · 15 · 18 Answer Given 3 . 6 . 9 . 12 . 15 . 18 Find: Convert into Factorial = (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6) = 3 6 (1 × 2 × 3 × 4 × 5 × 6) = 3 6 (6!) 4 C. Question Convert the following products into factorials : (n + 1)(n + 2) (n + 3) …(2n) Answer Given (n + 1)(n + 2) (n + 3) …(2n) Find: Convert into Factorial 4 D. Question Convert the following products into factorials : 1 · 3 · 5 · 7 · 9 …(2n – 1) Answer Given 1 · 3 · 5 · 7 · 9 …(2n – 1) Find: Convert into Factorial = 5 A. Question Which of the following are true : (2 + 3)! = 2! + 3! Answer Given: (2 + 3)! = 2! + 3! L.H.S = (2 + 3)! = 5! R.H.S = 2! + 3! = (2 × 1) + (3 × 2 × 1) = 2 + 6 = 8 Hence, L.H.S ? R.H.S 5 B. Question Which of the following are true : (2 × 3)! = 2! × 3! Answer Given: (2 × 3)! = 2! × 3! L.H.S = (2 × 3)! = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 R.H.S = 2! × 3! = 2 × 1 × 3 × 2 × 1 = 12 Hence, L.H.S ? R.H.S 6. Question Prove that: n! (n + 2) = n! + (n + 1)! Answer Given: n! (n + 2) = n! + (n + 1)! R.H.S. = n! + (n + 1)! = n! + (n + 1)(n + 1 – 1)! = n! + (n + 1)n! = n!(1 + n + 1) = n! (n + 2) = L.H.S L.H.S = R.H.S Hence, Proved.Read More
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