What is (n!)/((n-3)!+(n-2)!) ?

The solution to (n!)/((n-3)!+(n-2)!) is (n-2)!=(n-2)*(n-3)*\ldots*2*1

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(n!)/((n-3)!+(n-2)!)

\frac{n !}{( n - 3 ) ! + ( n - 2 ) !}

(n - 2)! = (n - 2) \cdot (n - 3) \cdot \dots \cdot 2 \cdot 1

Solution steps

Apply factorial rule:

n! = n \cdot (n - 1) \dots \cdot 2 \cdot 1

n! = n \cdot (n - 1) \cdot \dots \cdot 2 \cdot 1

Apply factorial rule:

n! = n \cdot (n - 1) \dots \cdot 2 \cdot 1

(n - 3)! = (n - 3) \cdot (n - 4) \cdot \dots \cdot 2 \cdot 1

Apply factorial rule:

n! = n \cdot (n - 1) \dots \cdot 2 \cdot 1

(n - 2)! = (n - 2) \cdot (n - 3) \cdot \dots \cdot 2 \cdot 1

4 x (2 x + 1) + 1 (2 x + 1)

\frac{36}{169}

\frac{4 - 3 x}{11} = 2

x (x - 4) \leq 12

5 b + 2 = 17

What is (n!)/((n-3)!+(n-2)!) ?
The solution to (n!)/((n-3)!+(n-2)!) is (n-2)!=(n-2)*(n-3)*\ldots*2*1