Now that you know how factorials work, you'll need to be able to simplify algebraic expressions and solve equations that use factorials in general form.

For example, if:

5! = 5 · 4 · 3 · 2 · 1

then the factorial of any number 'n' can be expressed as:

n! = n·(n - 1)·(n - 2)·(n - 3)·(n - 4)· ... ·1

Whatever expression is the starting point for a factorial results in an expression where each following term is one less:

(n + 3)! = (n + 3)·(n + 2)·(n + 1)·n·(n - 1)·(n - 2)·(n - 3)· ... ·1


Now have a look at some numerical examples where the fractional form means we can reduce:

6!  =  6·5·4·3·2·1  =  6·5·4·3·2·1  =  6·5  =  30
4!        4·3·2·1              4·3·2·1


8!  =   8·7·6·5·4·3·2·1  =  8·7·6·5·4·3·2·1  =  8·7·6  =  336
5!          5·4·3·2·1                 5·4·3·2·1   


Can you see that the need to write out the full factorial form, top and bottom, is not necessary, because the bottom will reduce with an identical sequence on top:

7!  =  7·6  =  42
5!


11!  =  11
10!


Here's one with algebraic terms, showing the full expansions:

   n!      =  n·(n - 1)·(n - 2)·(n - 3)· ... ·1  =  n·(n - 1)·(n - 2)·(n - 3)· ... ·1  =  n(n - 1)   = n2 - n
(n - 2)!         (n - 2)·(n - 3)· ... ·1                      (n - 2)·(n - 3)· ... ·1


The terms that are left over can often be simplified, here shown in grey.

(n+3)!  =   (n+3)(n+2)  =  n2 + 5n + 6
(n+1)!


(n + 1)!  =  (n + 1)·(n)·(n - 1)  =  (n + 1)·(n2 - n)  =  n3 - n2 + n2 - n  =  n3 - n
(n - 2)!



Equation Solving

example 1:

    n!      =   2
(n - 2)!


Start by simplifying the factorial expression:

  n·(n - 1)  =  2

Now solve the quadratic equation

  n2 - n = 2
  n2 - n - 2 = 0
  (n - 2)(n + 1) = 0
  n = 2    or   n = -1
  Because factorials are defined for Whole numbers only,
  the answer -1 is not permitted.
  This means n = 2 is the sole solution.

    Check:
        n!      =   2
    (n - 2)!
        2!      =   2
    (2 - 2)!
     2·1   =   2    [0! is defined as 1]
      1
    2  =  2



example 2:

    n!      =   6n
(n - 3)!


  n·(n - 1)·(n - 2) = 6n
  n(n2 - 3n + 2) = 6n
  n3 - 3n2 + 2n = 6n
  n3 - 3n2 - 4n = 0
  n·(n2 - 3n - 4) = 0
  n·(n - 4)·(n + 1) = 0
  n = 0   n = 4   n = -1
  Both n = 0 and n = -1 would result in factorials of negative numbers
  so they are not allowed.
  n = 4 is the sole answer.

    Check:
        n!      =   6n
    (n - 3)!
        4!      =   6(4)
    (4 - 3)!
    4·3·2·1 = 24
        1
    24 = 24



example 3:

2(n + 3)!  =  180
 (n + 1)!


  2(n + 3)(n + 2) = 180
  (n + 3)(n + 2) = 90
  n2 + 5n + 6 = 90
  n2 + 5n - 84 = 0
  (n + 12)(n - 7) = 0
  n = -12    n = 7
  Once again -12 won't work
  Solution n = 7

    Check:
    2(n + 3)!  =  180
     (n + 1)!
    2(7 + 3)!  =  180
     (7 + 1)!
    2(10)!  =  180
     (8)!
    2(10)(9) = 180
    180 = 180



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