# 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 ⋅ 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!)

=  (⋅ 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

(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

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### SAT Math Videos

May 22, 24 06:32 AM

SAT Math Videos (Part 1 - No Calculator)

2. ### Simplifying Algebraic Expressions with Fractional Coefficients

May 17, 24 08:12 AM

Simplifying Algebraic Expressions with Fractional Coefficients