HOW TO WRITE THE GIVEN PORDUCT OF NUMBERS AS FACTORIAL OF A NUMBER

Express in factorial form:

Problem 1 :

7 × 6 × 5

Solution :

7 × 6 × 5

Multiplying and dividing by 4!, we get

7! / 4! = (7 × 6 × 5) (4! / 4!)

Factorial form:

= 7! / 4!

Problem 2 :

10 × 9

Solution :

10 × 9

By decomposing 10, we get

10 = 2 x 5

Decomposing 9, we get

9 = 3 x 3

10 x 9 = 2 x 5 x 3 x 3

We cannot write it as factorial of some numerical value. So,

multiply the numerator and denominator by 8!, we get

10 × 9 = (10 x 9 x 8!)/8!

Factorial form:

10 × 9 = 10! / 8!

Problem 3 :

11 × 10 × 9 × 8 × 7

Solution :

Multiplying by 6!, we get

11! / 6! = (11 × 10 × 9 × 8 × 7) × (6! / 6!)

Factorial form:

11 × 10 × 9 × 8 × 7 = 11! / 6!

Problem 4 :

(13 × 12 × 11) / (3 × 2 × 1)

Solution:

13! / (10! 3!) = (13 × 12 × 11 × 10!) / 10! × (3 × 2 × 1)

Factorial form:    

(13 × 12 × 11) / (3 × 2 × 1) = 13! / (10! × 3!)

Problem 5 :

1 / (6 × 5 × 4)

Solution :

Multiplying by 3!/3!, we get

3! / 6! = 3! / (6 × 5 × 4 × 3!)

= 1/ (6 × 5 × 4)

Factorial form:

1/ (6 × 5 × 4) = 3! / 6!

Problem 6 :

(4 × 3 × 2 × 1) / (20 × 19 × 18 × 17)

Solution :

4! 16! / 20! = (4 × 3 × 2 × 1) × 16! / (20 × 19 × 18 × 17 × 16!)

Factorial form:    

(4 × 3 × 2 × 1) / (20 × 19 × 18 × 17) = (4! 16!) / 20!

Write each long expression as a single factorial.

Problem 7 :

1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ∙ 7 ∙ 8

Solution :

1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ∙ 7 ∙ 8 = 8!

Problem 8 :

1 ∙ 2 ∙ 3……100

Solution :

1 ∙ 2 ∙ 3……100 = 100!

Problem 9 :

1 ∙ 2 ∙ 3 ∙ 4 …. (n - 2) ∙ (n - 1) ∙ n

Solution :

1 ∙ 2 ∙ 3 ∙ 4 …. (n - 2) ∙ (n - 1) ∙ n = n!

Problem 10 :

1 ∙ 2 ∙ 3 ∙ …. 56 ∙ 57 ∙ 58

Solution :

1 ∙ 2 ∙ 3 ∙ …. 56 ∙ 57 ∙ 58 = 58!

Problem 11 :

1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ … (5n - 4) ∙ (5n - 3) ∙ (5n - 2) ∙ (5n - 1)

Solution :

1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ … (5n - 4) ∙ (5n - 3) ∙ (5n - 2) ∙ (5n - 1)

Considering the given expansion in reverse form, we get

= (5n - 1) ∙ (5n - 2) ∙ (5n - 3) ........5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

= (5n - 1)!