FORMULA USED: n!= n(n-1)(n-2)...(3)(2)(1), where n is an element of the positive integers.
Example 4:
5!= ----->n=5 so when you plug in to the formula you get:
5(5-1)(5-2)(5-3)(5-4)
(5)(4)(3)(2)(1)
=120
We can remember these factorials:
- 0!= 1
- 1!= 1
- 2!= 2
- 3!= 6
- 4!= 24
- 5!= 120
- 6!= 720
- 7!= 5040
- 8!= 40320
- 9!= 362880
- 10!= 3628800
We can also simplify factorial equations:
Example 5:
a) 5!/4! ----->we can expand the 5! to (5)(4!)
(5)(4!)/4! ----->and the (4!)'s are able to cancel out each other leaving the (5)
=5
e) (s-2)!/(s+1)! ----->we can expand the (s+1)! into: (s+1)(s+0)(s-1)(s-2)!
=1/(s+1)(s+0)(s-1)
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