Solve 8!/3!+5-9!/2!-7!

$$\frac { 8 ! } { 3 ! + 5 } - \frac { 9 ! } { 2 ! - 7 ! }$$

Evaluate

$$\frac{9414720}{2519}\approx 3737.483128225$$

Solution Steps

The factorial of 8 is 40320.

$$\frac{40320}{3!+5}-\frac{9!}{2!-7!}$$

The factorial of 3 is 6.

$$\frac{40320}{6+5}-\frac{9!}{2!-7!}$$

Add 6 and 5 to get 11.

$$\frac{40320}{11}-\frac{9!}{2!-7!}$$

The factorial of 9 is 362880.

$$\frac{40320}{11}-\frac{362880}{2!-7!}$$

The factorial of 2 is 2.

$$\frac{40320}{11}-\frac{362880}{2-7!}$$

The factorial of 7 is 5040.

$$\frac{40320}{11}-\frac{362880}{2-5040}$$

Subtract 5040 from 2 to get -5038.

$$\frac{40320}{11}-\frac{362880}{-5038}$$

Reduce the fraction \frac{362880}{-5038} to lowest terms by extracting and canceling out 2.

$$\frac{40320}{11}-\left(-\frac{181440}{2519}\right)$$

The opposite of -\frac{181440}{2519} is \frac{181440}{2519}.

$$\frac{40320}{11}+\frac{181440}{2519}$$

Least common multiple of 11 and 2519 is 2519. Convert \frac{40320}{11} and \frac{181440}{2519} to fractions with denominator 2519.

$$\frac{9233280}{2519}+\frac{181440}{2519}$$

Since \frac{9233280}{2519} and \frac{181440}{2519} have the same denominator, add them by adding their numerators.

$$\frac{9233280+181440}{2519}$$

Add 9233280 and 181440 to get 9414720.

$$\frac{9414720}{2519}$$

Factor

$$\frac{2 ^ {6} \cdot 3 ^ {2} \cdot 5 \cdot 7 \cdot 467}{11 \cdot 229} = 3737\frac{1217}{2519} = 3737.4831282254863$$