Factorial Question

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  • #1
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How do you get

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

????

Isn't (n + 1)! = (n + 1) ⋅ (n + 1 - 1) ⋅ (n + 1 - 2) ⋅ (n + 1 - 3) ⋅ (n + 1 - 4) ... and so on?
 

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  • #2
Simon Bridge
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How do you get

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

????

Isn't (n + 1)! = (n + 1) ⋅ (n + 1 - 1) ⋅ (n + 1 - 2) ⋅ (n + 1 - 3) ⋅ (n + 1 - 4) ... and so on?
You need to realise that:
n+1-1 = n
n+1-2 = n-1
n+1-3 = n-2
etc
 
  • #3
12,209
5,902
or more compactly:

(n+1)! = (n+1) * (n+1-1)! = (n+1) * n!
 

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