Simplify the factorials in each fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form.

$\displaystyle \frac{2!}{13!}\div \frac{3!}{15!}$

$\displaystyle \frac{2\cdot 1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\div \frac{3\cdot 2\cdot 1}{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 }$

$\displaystyle \frac{1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}\div \frac{1}{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$

Since it is division of fractions change the division sign to multiplication and flip the second fraction

$\displaystyle \frac{1}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}\cdot \frac{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{1}$

$\displaystyle \frac{15\cdot 14\cdot13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}$

$\displaystyle \frac{15\cdot 14}{3}$

$\displaystyle \frac{210}{3}$

$\displaystyle 70$

Simplify the factorials in the fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form, but I would suggest cancelling out the 7! in the numerator and denominator first.

$\displaystyle \frac{8!\cdot 7!\cdot 3!}{9!\cdot 7!}$

$\displaystyle \frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1}{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$\displaystyle \frac{3\cdot 2\cdot 1}{9}$

$\displaystyle \frac{6}{9}$

Then reduce the fraction to lowest terms

$\displaystyle \frac{2}{3}$

Simplify the factorials in the fraction by canceling common factors in the numerator and denominator. It can help to write it in expanded form.

$\displaystyle \frac{4!}{6!}\cdot \frac{7!}{2!}\cdot 2!$

$\displaystyle \frac{4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\cdot \frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1}\cdot 2\cdot 1$

$\displaystyle \frac{1}{6\cdot 5}\cdot \frac{7\cdot 6\cdot 5\cdot 4\cdot 3}{1}\cdot 2\cdot 1$

Combine each of the fractions into one fraction.

$\displaystyle \frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5}$

$\displaystyle \frac{7\cdot 4\cdot 3\cdot 2\cdot 1}{1}$

$\displaystyle \frac{168}{1}$

$\displaystyle 168$

Dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

$\displaystyle \frac{3!}{5!} \div \frac{5!}{3!}=\frac{3!}{5!}\times \frac{3!}{5!}$

Write out the terms of the factorials.

$\displaystyle \frac{3!}{5!}\times \frac{3!}{5!}= \frac{3\times 2\times 1}{5\times 4\times 3\times 2\times 1}\times \frac{3\times 2\times 1}{5\times 4\times 3\times 2\times 1}$

Cancel all common terms.

$\displaystyle \frac{1}{5\times 4}\times\frac{1}{5\times 4} = \frac{1}{20}\times \frac{1}{20} = \frac{1}{400}$

The answer is:  $\displaystyle \frac{1}{400}$

Evaluate the terms in parentheses first.  Do not distribute the integer through the parentheses or this will change the value of the factorial!

$\displaystyle \frac{2(4+1)!}{3!(2+3)!} =\frac{2(5)!}{3!(5)!}$

The common terms in the numerator and denominator can be simplified.

$\displaystyle \frac{2}{3!} = \frac{2}{3 \times 2\times 1} = \frac{1}{3}$

The answer is:  $\displaystyle \frac{1}{3}$

Simplify the factorials first.

$\displaystyle 3! = 3\cdot2\cdot1 = 6$

$\displaystyle 4! = 4\cdot3\cdot2\cdot1 = 24$

Replace the values back into the problem.

$\displaystyle \frac{3(2+3!)}{1-4!} = \frac{3(2+6)}{1-24} = \frac{24}{-23}$

The answer is:  $\displaystyle -\frac{24}{23}$

Simplify the denominator.

$\displaystyle \frac{2(3!+4)}{(2+2)!}=\frac{2(3!+4)}{4!}$

Rewrite each factorial in the given fraction.  Do not distribute the two through the quantity or this will change the value of the factorial.

$\displaystyle \frac{2(3!+4)}{4!} =\frac{2([3\times 2\times 1]+4)}{[4\times3\times 2\times 1] } = \frac{2(6+4)}{24}=\frac{2(10)}{24}=\frac{20}{24}$

Reduce this fraction.

The answer is:  $\displaystyle \frac{5}{6}$

To simplify this, we will need to evaluate the parentheses first.  Do not distribute the integer two into the binomial, or the factorial value will change.

$\displaystyle \frac{2(3+2)!}{3!}=\frac{2(5)!}{3!}$

Expand the factorials.

$\displaystyle \frac{2(5)!}{3!} = \frac{2(5\cdot 4\cdot 3\cdot 2\cdot 1)}{3\cdot 2\cdot 1}$

Simplify the top and bottom terms.

$\displaystyle 2(5\cdot 4) = 40$

The answer is:  $\displaystyle 40$

Simplify the inner parentheses and expand the factorials.

$\displaystyle (3!)(3+2)!(\frac{1}{4!}) = (3\times 2\times 1)(5)!(\frac{1}{4 \times 3\times 2\times 1})$

Cancel out common terms.

$\displaystyle (5\times 4\times 3\times 2\times 1)(\frac{1}{4}) = 5\times 3\times 2\times 1$

The answer is:  $\displaystyle 30$

In order to solve this, we will need to expand all the factorials.

$\displaystyle \frac{2!}{3!}\cdot \frac{4!}{6!} = \frac{2\times1}{3\times2\times1}\cdot \frac{4\times3\times2\times1}{6\times5\times4\times3\times2\times1 }$

Cancel all the common terms and write the remaining numbers.

$\displaystyle \frac{1}{3 }\cdot \frac{1}{30} = \frac{1}{90}$

The answer is:  $\displaystyle \frac{1}{90}$