The coefficient of the constant term is $\left( \begin{matrix} 2014 \\ 1007 \end{matrix} \right) = \dfrac {2014!} {1007!^2}$

The number of trialing zeros of $n!$ is given by: $\displaystyle z = \sum _{i=1}^{k} \left \lfloor \dfrac {n}{5^i} \right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function .

$\begin{aligned} \implies z(2014) & = \left \lfloor \frac {2014}{5} \right \rfloor + \left \lfloor \frac {2014}{25} \right \rfloor + \left \lfloor \frac {2014}{125} \right \rfloor + \left \lfloor \frac {2014}{625} \right \rfloor \\ & = 402 + 80 + 16 + 3 = 501 \\ \implies z(1007) & = \left \lfloor \dfrac {1007}{5} \right \rfloor +\left \lfloor \dfrac {1007}{25} \right \rfloor +\left \lfloor \dfrac {1007}{125} \right \rfloor + \left \lfloor \dfrac {1007}{625} \right \rfloor \\ & = 201+40+8+1=250 \end{aligned}$

The trialing zeros of $\dfrac {2014!} {1007!^2}$ , $z = z(2014) - 2z (1007) = 501 - 500 = \boxed{1}$

This is going to be a long ride,so sit back and enjoy. Let us start by expanding $(x+\dfrac{1}{x})^{2014}$ .We will do this using the binomial theorem: $(x+\dfrac{1}{x})^{2014}=\dbinom{2014}{0}*x^{2014}*(\dfrac{1}{x})^{0}+\dbinom{2014}{1}*x^{2013}*(\dfrac{1}{x})^{1}\\....\dbinom{2014}{2013}*x^{1}*(\dfrac{1}{x})^{2013}+\dbinom{2014}{2014}*x^{0}*(\dfrac{1}{x})^{2014}.$ Now,because we want a constant term there shouldn't be any $x$ in it.Now,because $x$ and $\dfrac{1}{x}$ are reciprocals they will cancel each other but in the expansion above $x$ and $\dfrac{1}{x}$ have the same power only once and that is $1007$ .The term is: $\dbinom{2014}{1007}*x^{1007}*(\dfrac{1}{x})^{1007}=\dfrac{2014!}{1007!*1007!}*1.$ Now,find the number of trailing zeroes in the numerator and denominator and we are done.How to find the number of zeroes at the end of $n!:$ $[\dfrac{n}{5}]+[\dfrac{n}{5^{2}}]+......$ Till $5^{x}<n.$ Where[] is the greatest integer function.

Constant term wil be 2014 C 1007, 1007 will be derived from laws of exponent, x^(2014-T)*x^(-T) = x^(0), where, 2014-T+(-T) = 0, 2T=2014, then T=2014/2 = 1007. Next, Trailing zero's: Count numbers within factorial divisible by 5, 25, 125, 625 and so on (base 5 divisors):

Consider 2014, divisor 5, max number near 2014 = 2010, so 2010/5= 402 + divisor 25, max number near 2014 = 2000, so 2000/25= 80 + divisor 125, max number near 2014 = 2000, so 2010/125= 16 + divisor 625, max number near 2014 = 1875, so 1875/625= 3 , divisor 3125 > 2014 , so stop. Total No. of trailing zero's = 402 + 80 + 16 + 3 = 501.

Consider 1007, divisor 5, max number near 1007= 1005, so 1005/5= 201+ divisor 25, max number near 1007= 1000, so 1000/25= 40+ divisor 125, max number near 1007= 1000, so 1000/125= 8+ divisor 625, max number near 1007 = 625, so 625/625= 1, divisor 3125 > 1007, so stop. Total No. of trailing zero's = 201+ 40+ 8+ 1 = 250.

Going Back to Constant= 2014C1007 = 2014! / [ (2014-1007)! 1007! ] =2014! / ( 1007! 1007! ) = 2014! / (1007! )^2

of trailing zeros, Numerator - Denominator = 501 - 250 * 2= 501-500 = 1

Applying de Polignac's formula helps.

I did it like this: $\newline$ The coefficient of the constant term is $\begin{aligned} {2014 \choose 1007} = \frac{2014!} {1007!^2} &= \frac{2014 \cdot 2013 \cdot ... \cdot 1008} {1007!} \end{aligned} \newline$ After that I looked at the numerator and denominator separately and calculated how many factors with trailing zeroes they have, since these are also the only reason for trailing zeroes in the result. For $2014 \cdot 2013 \cdot ... \cdot 1008$ we can examine the interval $1000$ to $2010$ and count how many $10$ ths, $100$ ths, etc. there are, which also represents the amount of multiplicands with one, two or three zeroes at the end. Doing that, we get $\frac{1010}{10} = 101 \;10$ ths, $\frac{1010}{100} = 10 \;100$ ths, and $\frac{1010}{1000} = 1 \;1000$ ths. Summing that up we get $121$ zeroes. Analogously with the interval $1000$ to $2000$ we get $120$ zeroes, meaning we get something like $\frac{X \cdot 10^{121}}{Y \cdot 10^{120}}$ leaving us with only one factor of $10$ and thus one trailing zero. $\newline$ But I am not really sure if I explained it well or if this is even a valid solution method.

The expression that you must find the number of trailing 0s is c=a/b where a is the product of all x: 1007<x<=2014 and b is the product of all x: 0<x<=1007, you than take the prime factorisation of a and b, and find the difference in the exponents for 5 in a and 5 in b, let us call that i, and than find the difference in the exponents for 2 in a and b, let us call that j. choose either i or j (just take the lower one). I technacly cheated by first trying out a few assumptions as I was too lazy to do the whole method first time, I assumed j to be higher than i so I could just use i.

We have to find {Trailing zeros in 2014! - Trailing zeros in (1007!)^2}

=Trailing Zeros in (1008.1009.....2014) - Trailing zeros in (1.2.3.......1007)

=Trailing Zeros in (8.9..........1014) - Trailing zeros in (1.2.3.......1007) =Trailing Zeros in (1008.1009...1014) =Trailing Zeros in 1010 =1