When evaluating a factorial, you multiply the original number by each integer less than it, stopping at 1.

For this problem, this means that

 \(\displaystyle 5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 120\).

Then adding 3, we get the answer 123

The value of \(\displaystyle e\) is defined as \(\displaystyle 2.71828\).

To find \(\displaystyle e^3\), simply cube the decimal number.

\(\displaystyle e^3=2.71828\times 2.71828\times 2.71828 = 20.085\)

The closest value of this number is:  \(\displaystyle 20\)

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

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.