Solve 1/5!+1/3!+1/3

$$\frac { 1 } { 5 ! } + \frac { 1 } { 3 ! } + \frac { 1 } { 3 }$$

Evaluate

$$\frac{61}{120}\approx 0.508333333$$

Solution Steps

The factorial of 5 is 120.

$$\frac{1}{120}+\frac{1}{3!}+\frac{1}{3}$$

The factorial of 3 is 6.

$$\frac{1}{120}+\frac{1}{6}+\frac{1}{3}$$

Least common multiple of 120 and 6 is 120. Convert \frac{1}{120} and \frac{1}{6} to fractions with denominator 120.

$$\frac{1}{120}+\frac{20}{120}+\frac{1}{3}$$

Since \frac{1}{120} and \frac{20}{120} have the same denominator, add them by adding their numerators.

$$\frac{1+20}{120}+\frac{1}{3}$$

Add 1 and 20 to get 21.

$$\frac{21}{120}+\frac{1}{3}$$

Reduce the fraction \frac{21}{120} to lowest terms by extracting and canceling out 3.

$$\frac{7}{40}+\frac{1}{3}$$

Least common multiple of 40 and 3 is 120. Convert \frac{7}{40} and \frac{1}{3} to fractions with denominator 120.

$$\frac{21}{120}+\frac{40}{120}$$

Since \frac{21}{120} and \frac{40}{120} have the same denominator, add them by adding their numerators.

$$\frac{21+40}{120}$$

Add 21 and 40 to get 61.

$$\frac{61}{120}$$

Factor

$$\frac{61}{2 ^ {3} \cdot 3 \cdot 5} = 0.5083333333333333$$