SOLVING FACTORIAL PROBLEMS

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 ⋅ ⋅ 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!)

  =  (⋅ 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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