$\displaystyle \frac{8!}{5!}=\frac{8*7*6*5*4*3*2*1}{5*4*3*2*1}$

The 5,4,3,2,1 cancel out leaving 8*7*6 which is 336

Rewrite the factorials.

$\displaystyle 4! = 4\times3\times2\times1 = 24$

$\displaystyle 3!=3\times2\times1 = 6$

Multiply these two numbers together.

$\displaystyle 24\times 6 = 144$

The answer is:  $\displaystyle 144$

Expand both factorials in the numerator and denominator.

$\displaystyle \frac{6!}{3!} = \frac{6 \times 5\times4\times3\times2\times1}{3\times2\times1}$

Cancel the common terms on the numerator and denominator.

The numerator becomes:  

$\displaystyle 6\times 5 \times 4 =120$

The answer is:  $\displaystyle 120$

Do not add the factorials!

$\displaystyle 3!+2! \neq 5!$

We will need to simplify and rewrite the terms of the factorials.

$\displaystyle \frac{3!(3!+2!)^2}{5!} = \frac{(3\times 2\times 1)[(3\times 2\times 1)+(2\times 1)]^2}{5\times 4\times3\times 2\times 1}$

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

$\displaystyle \frac{[(3\times 2\times 1)+(2\times 1)]^2}{5\times 4}$

Simplify the numerator and denominator.

$\displaystyle \frac{[6+2]^2}{20} = \frac{8^2}{20} = \frac{64}{20} =\frac{4\times 4\times 4}{5\times 4}$

Simplify the fraction by cancelling the common terms.

The answer is:  $\displaystyle \frac{16}{5}$

Rewrite all the factorial terms in expanded form.  To expand a factorial, multiply all the integers in decreasing order until it reaches to one.

$\displaystyle 3!(\frac{(4+3)!}{7(5!)}) = (3\times 2\times 1)(\frac{7!}{7(5!)}) = 6(\frac{7!}{7(5!)})$

Do not distribute the integer through the number before the factorial, and do not cancel the sevens in the numerator and denominator.

$\displaystyle 7(5!) \neq 35!$

$\displaystyle \frac{7!}{7} \neq 1!$

Rewrite the numerator in terms so that we can simplify the fraction.

$\displaystyle 6(\frac{7!}{7(5!)}) = 6(\frac{7\times 6\times5\times4\times3\times2\times1}{7(5\times4\times3\times2\times1)})$

Notice all the constants that we can cancel.  This fraction will reduce to"

$\displaystyle 6\times 6 = 36$

The answer is:  $\displaystyle 36$

Expand the factorials.  

$\displaystyle 3!\times (2+3)!\times 0! = (3\times 2\times 1)\times 5! \times 0!$

$\displaystyle =(3\times 2\times 1)\times(5\times 4\times 3\times 2\times 1)\times 0!$

Zero factorial is a special case.  It is equal to one.

$\displaystyle 0!=1$

The expression becomes:

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

Multiply all the numbers.

$\displaystyle =6\times 120= 720$

The answer is:  $\displaystyle 720$

Simplify the factorials in the numerator first.  

$\displaystyle 2!(3+3)! = 2!\cdot 6! =(2\times 1)\cdot (6\times5\times4\times3\times2\times1)$

Simplify the denominator.

$\displaystyle 3!\times 6 = (3\times 2\times 1) \times 6$

Eliminate common terms in the numerator and denominator.

The numerator becomes:  $\displaystyle 2 \times 1\times5\times4 = 40$

The answer is:  $\displaystyle 40$

Evaluate the first term.  Write out the terms of the factorial.

$\displaystyle (7-4)! = 3! = 3\times 2 \times 1 = 6$

Evaluate the second term.

$\displaystyle (7!-5!)$

$\displaystyle 7! = 7\times6\times5\times4\times3\times2\times1 = 5040$

$\displaystyle 5!=5\times4\times3\times2\times1 = 120$

$\displaystyle (7!-5!) = 5040-120 = 4920$

Combine the simplified terms.

$\displaystyle (7-4)!\cdot(7!-5!) = 6\cdot 4920$

The answer is:  $\displaystyle 29520$

When you have factorials in the numerator and denominator of a fraction, you can cancel out the common factors between them. It can help to write the problems in expanded form.

$\displaystyle \frac{15!}{13!}$

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

The entire denominator 13 will be cancelled out leaving only

$\displaystyle 15\cdot 14$

$\displaystyle 210$

Similar factors that are in both the numerator and denominator of a fraction cancel out. It helps to write it in expanded form.

$\displaystyle \frac{7!}{4!}\cdot \frac{5!}{2!}$

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

$\displaystyle 7\cdot 6\cdot 5 \cdot 5\cdot 4\cdot 3$

$\displaystyle 12600$