\(\displaystyle 5!/3! = (5*4*3*2*1)/(3*2*1)\)

Simplifying this equation we notice that the 3's, 2's, and 1's cancel so

\(\displaystyle 5!/3! = 5*4 = 20\)

Alternative Solution

\(\displaystyle 5!/3! = (120/6) = 20\)

The \(\displaystyle 4!\) cancels out all of \(\displaystyle 6!\) except for the parts higher than 4, this leaves a 6 and a 5 left to multilpy \(\displaystyle 5!\)

Recall that \(\displaystyle (n+3)! = (n+3)(n+2)(n+1)(n)...\).

Likewise, \(\displaystyle 6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\).

Thus, the expression \(\displaystyle \frac{7!(n+3)!}{6!(n+2)!}\) can be simplified in two parts:

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

and

\(\displaystyle \frac{(n+3)!}{(n+2)!} = \frac{(n+3)(n+2)(n+1...)}{(n+2)(n+1)...} = n+3\)

The product of these two expressions is the final answer: \(\displaystyle 7n + 21\)

To simplify this, just write out each factorial:

\(\displaystyle \frac{6!}{2!4!} = \frac{6*5*4*3*2*1}{(2*1)(4*3*2*1)} = \frac{6*5}{2*1} = 15\)

The factorial sign (!) just tells us to multiply that number by every integer that leads up to it.  So, \(\displaystyle \frac{7!}{3!}\) can also be written as: 

\(\displaystyle \frac{(7\times6\times5\times4\times3\times2\times1)}{(3\times2\times1)}\)

To make this easier for ourselves, we can cancel out the numbers that appear on both the top and bottom:

\(\displaystyle 7\times6\times5\times4 = 840\)

This is a factorial question. The formula for factorials is \(\displaystyle n! = n\times(n-1)\times(n-2)\times...\times1\).

 \(\displaystyle 6!=6\times5\times4! = 30\times 4!\)

A factorial is a number which is the product of itself and all integers before it. For example \(\displaystyle 5!=5\cdot 4 \cdot 3 \cdot 2 \cdot 1\)

In our case we are asked to divide \(\displaystyle 2015!\) by \(\displaystyle 2014!\). To do this we will set up the following:

\(\displaystyle \frac{2015!}{2014!}\)

We know that \(\displaystyle 2015!\) can be rewritten as the product of itself and all integers before it or:

\(\displaystyle 2015!=2015 \cdot 2014!\)

Substituting this equivalency in and simplifying the term, we get:

\(\displaystyle \frac{2015 \cdot 2014!}{2014!}=2015\)

This expression of factorials reduces to (n+1)(n+2).  Therefore, the solution must be a number that multiplies to 2 consecutive integers.  Only 30 is a product of 2 consecutive integers.  \(\displaystyle 5\ast6=30\)

So n would have to be 4 in this problem.

Remember what a factorial is, and first write out what the original equation means. A factorial is a number that you multiply by all whole numbers that come before it until you reach one.

You can simplify because all terms in the expression 17! are found in 20!.

Thus:

\(\displaystyle \frac{20!}{17!\times 2!}=\frac{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot 2}\)

\(\displaystyle =\frac{20\cdot19\cdot18}{2}=10\cdot19\cdot18=3420\)

Rewrite the factorials in multiplicative order.

\(\displaystyle \frac{4!+1}{3!+1} = \frac{(4\times 3\times2\times1)+1}{(3\times2\times1)+1}\)

In this scenario, the numbers of the factorial in the numerator and denominator CANNOT cancel.  Simplify by multiplying out the factorials.

\(\displaystyle \frac{(4\times 3\times2\times1)+1}{(3\times2\times1)+1}= \frac{24+1}{6+1}=\frac{25}{7}\)