Solve 5!/5-3!*3!*(4/1024)=90/1024

$$\frac { 5 ! } { 5 - 3 ! \times 3 ! } \times ( \frac { 4 } { 1024 } ) = \frac { 90 } { 1024 }$$

Verify

Solution Steps

Multiply 3! and 3! to get \left(3!\right)^{2}.

$$\frac{5!}{5-\left(3!\right)^{2}}\times \frac{4}{1024}=\frac{90}{1024}$$

The factorial of 5 is 120.

$$\frac{120}{5-\left(3!\right)^{2}}\times \frac{4}{1024}=\frac{90}{1024}$$

The factorial of 3 is 6.

$$\frac{120}{5-6^{2}}\times \frac{4}{1024}=\frac{90}{1024}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{120}{5-36}\times \frac{4}{1024}=\frac{90}{1024}$$

Subtract 36 from 5 to get -31.

$$\frac{120}{-31}\times \frac{4}{1024}=\frac{90}{1024}$$

Fraction \frac{120}{-31} can be rewritten as -\frac{120}{31} by extracting the negative sign.

$$-\frac{120}{31}\times \frac{4}{1024}=\frac{90}{1024}$$

Reduce the fraction \frac{4}{1024} to lowest terms by extracting and canceling out 4.

$$-\frac{120}{31}\times \frac{1}{256}=\frac{90}{1024}$$

Multiply -\frac{120}{31} times \frac{1}{256} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-120}{31\times 256}=\frac{90}{1024}$$

Do the multiplications in the fraction \frac{-120}{31\times 256}.

$$\frac{-120}{7936}=\frac{90}{1024}$$

Reduce the fraction \frac{-120}{7936} to lowest terms by extracting and canceling out 8.

$$-\frac{15}{992}=\frac{90}{1024}$$

Reduce the fraction \frac{90}{1024} to lowest terms by extracting and canceling out 2.

$$-\frac{15}{992}=\frac{45}{512}$$

Least common multiple of 992 and 512 is 15872. Convert -\frac{15}{992} and \frac{45}{512} to fractions with denominator 15872.

$$-\frac{240}{15872}=\frac{1395}{15872}$$

Compare -\frac{240}{15872} and \frac{1395}{15872}.

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