Problem 1 :
Find the value of
(i) 6!
(ii) 4! + 5!
(iii) 3! − 2!
(iv) 3! × 4!
(v) 12! / (9! × 3!)
(vi) (n + 3)! / (n + 1)!
Solution :
(i) 6! = 6⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 |
(ii) 4! + 5! = 4! + 5 ⋅ 4! = 4! (1 + 5) = 24 (6) = 144 |
(iii) 3! − 2! = 3 ⋅2! - 2! = 2! (3 - 1) = 2 (2) = 4 |
(iv) 3! × 4! = (3 ⋅ 2 ⋅ 1) (4 ⋅ 3 ⋅ 2 ⋅ 1) = 6 ⋅ 24 = 144 |
(v) 12! / (9! × 3!) = (12⋅11⋅10⋅9!) /(9! ⋅ 3!) = (12⋅11⋅10)/(3⋅2⋅1) = 2 ⋅ 11 ⋅ 10 = 220 |
(vi) (n + 3)! / (n + 1)! = (n+3)(n+2)(n+1)!/(n+1)! = (n+3)(n+2) = n2 + 3n + 2n + 6 = n2 + 5n + 6 |
Problem 2 :
Evaluate n! / r!(n − r)! when
(i) n = 6, r = 2 (ii) n = 10, r = 3 (iii) For any n with r = 2.
Solution :
(i) n! / r!(n − r)!
= 6!/(2! ⋅ 4!)
= (6 ⋅ 5 ⋅ 4!)/(2! ⋅ 4!)
= (6 ⋅ 5) / 2
= 15
(ii) n = 10, r = 3
n! / r!(n − r)!
= 10!/(3! ⋅ 7!)
= (10 ⋅ 9 ⋅ 8 ⋅ 7!)/(3! ⋅ 7!)
= (10 ⋅ 3 ⋅ 4)
= 120
(iii) For any n with r = 2.
n! / r!(n − r)!
= n!/2!(n - 2)!
= [n(n - 1)(n-2)!] / [2!(n - 2)!]
= n (n - 1) / 2
Problem 3 :
Find the value of n if
(i) (n + 1)! = 20(n − 1)! (ii) (1/8!) + (1/9!) = (n/10!)
Solution :
(i) (n + 1)! = 20(n − 1)!
(n + 1) n (n - 1)! = 20 (n - 1)!
n(n + 1) = 20
n2 + n - 20 = 0
(n - 4)(n + 5) = 0
n = 4 and n = -5
Hence the answer is 4.
(ii) (1/8!) + (1/9!) = (n/10!)
(1/8!) + (1/9)(1/8!) = (n/90)(1/8!)
1 + (1/9) = n/90
10/9 = n/90
n = 100
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