Simplify the denominator in the parentheses first.

\(\displaystyle \frac{8!}{(2+4)!} = \frac{8!}{6!}\)

Rewrite the factorials.

\(\displaystyle \frac{8!}{6!} = \frac{8\times7\times6\times5\times4\times3\times2\times1 }{6\times5\times4\times3\times2\times1}\)

Cancel all the common terms in the numerator and denominator.

The numerator becomes:  \(\displaystyle 8\times 7 =56\)

The answer is:  \(\displaystyle 56\)

In order to simplify the factorials, first simplify the parentheses.

\(\displaystyle (1+2)! \cdot 3!= 3! \cdot 3!\)

Simplify the factorials.

\(\displaystyle (3\times 2 \times 1)\cdot(3\times 2 \times 1) = (6) \cdot (6) =36\)

The answer is:  \(\displaystyle 36\)

The value of zero factorial is one.

\(\displaystyle 0!=1\)

Replace this value.

\(\displaystyle \frac{(1+0!)!}{0!+3!}=\frac{(1+1!)!}{1+3!}\)

Expand the factorials.

\(\displaystyle \frac{(1+1!)!}{1+3!}=\frac{(1+1)!}{1+(3\times 2\times 1)}= \frac{2}{1+6}=\frac{2}{7}\)

The answer is:  \(\displaystyle \frac{2}{7}\)

Simplify the numerator of the fraction.

\(\displaystyle \frac{(3+5)!}{6!} = \frac{8!}{6!}\)

Do not reduce the fraction and cancel out the factorials.  Instead, expand both.

\(\displaystyle \frac{8!}{6!} = \frac{8\times7\times6\times5\times4\times3\times2\times1}{6\times5\times4\times3\times2\times1}\)

Notice that the similar terms on the numerator and denominator will cancel.

The remaining terms are:  \(\displaystyle 8\times 7= 56\)

The answer is:  \(\displaystyle 56\)

Simplify every term inside the parentheses.

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

Expand each factorial.

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

Cancel out the common terms on the numerator and denominator and rewrite what remains of the numerator.

\(\displaystyle 4\times 2\times 1=8\)

The answer is:  \(\displaystyle 8\)

Expand each factorial.

\(\displaystyle \frac{3! \cdot5!}{4!\cdot6!}=\frac{(3\times2\times1) (5\times4 \times3\times2\times1)}{ (4 \times3\times2\times1)(6\times 5\times4 \times3\times2\times1)}\)

Cancel out the common terms on the numerator and denominator.

The remain terms are:

\(\displaystyle \frac{1}{4(6)} = \frac{1}{24}\)

The answer is:  \(\displaystyle \frac{1}{24}\)

In order to divide these fractions, we will need to take the reciprocal of the second term and change the sign to a multiplication.

\(\displaystyle \frac{3}{8}\times \frac{34}{9}\)

Write each non-prime term in factors.

\(\displaystyle \frac{3}{2\times 4}\times \frac{2\times 17}{3 \times 3}\)

Reduce the terms by cancelling the common factors in the numerator and denominator.

\(\displaystyle \frac{17}{4\times 3} = \frac{17}{12}\)

The answer is:  \(\displaystyle \frac{17}{12}\)

Identify the values of each factorial.  Zero factorial is a special case.

\(\displaystyle 0!=1\)

\(\displaystyle 2! = 2\times 1 = 2\)

\(\displaystyle 4! =4\times 3 \times 2\times 1=24\)

Multiply all three numbers.

\(\displaystyle 0!\cdot 2! \cdot 4!=1\cdot2\cdot24 = 48\)

The answer is:  \(\displaystyle 48\)

Simplify the numerator.  Do not distribute the five through the terms inside the parentheses, or it will change the value of the factorial!

\(\displaystyle 5(3+2)! = 5(5)! = 5[5\cdot 4\cdot 3\cdot 2\cdot 1]\)

Simplify the denominator.

\(\displaystyle 5!\cdot 3! = [5\cdot 4\cdot 3\cdot 2\cdot 1]\cdot [3\cdot 2\cdot 1]\)

Divide the numerator with the denominator.

\(\displaystyle \frac{5[5\cdot 4\cdot 3\cdot 2\cdot 1]}{[5\cdot 4\cdot 3\cdot 2\cdot 1]\cdot [3\cdot 2\cdot 1]}= \frac{5}{3\cdot 2\cdot 1}= \frac{5}{6}\)

The answer is:  \(\displaystyle \frac{5}{6}\)

Simplify the terms in side the parentheses first.  Do NOT distribute the four through the binomial or this will change the value of the factorial!

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

Notice the there are common terms in the numerator and denominator.  Cancel out the terms.  The remaining parts are: 

\(\displaystyle \frac{4\cdot 4}{2} = 8\)

The answer is:  \(\displaystyle 8\)