Number of ways to get a total of 14 when tossing a die 3 times Each of the 3 boys tosses a die once. Find the number of ways for them to get a total of 14. 
I'm trying to solve it by forming this equation, 
$$x_1 + x_2 + x_3 = 14$$
where $1\leq x_i\leq 6$ for $i=1,2,3$. 
The answer given is 15 but I'm getting the number 33 trying to solve the equation. Where do I go wrong?
EDIT:
I did solve the question using generating function by finding the coefficient of $x^{14}$ in the function $(x+x^2+x^3+x^4+x^5+x^6)^3$. But I'm curious to why I'm getting the wrong answer when I'm trying to solve my equation that I formed. Any ideas?
How I solved my equation:
Find all the non-negative integer solutions of $x_1 + x_2 + x_3 = 11$ which is $\binom{11+3-1}{11}$ which is $\binom{13}{11}$.
I let $P_i$ be the property that $x_i \geq 7$.
$\omega(0) = |S| = \binom{13}{11}$
$\omega(1) = \omega(P_1) + \omega(P_2) + \omega(P_3) = 3 \binom{(11-7)+3-1}{(11-7)}$.
To find $E(0) = \omega(0) - \omega(1) + \omega(2) - \omega(3) = 33$
 A: Using your definitions of $x_i$, $1\leq x_3\leq 6$, so that the sum of the first two is any of
$$x_1+x_2=8,9,10,11,12,13$$
For $8$, we have $5$ combinations of $(x_1,x_2)$; for $9$ we have $4$; for $10$, we have $3$; $11$, $2$; $12$, $1$; $13$, $0$. Adding these up together, there are $5+4+3+2+1+0=15$ ways to get the three dice to add to $14$. This method can be generalised to higher amounts of dice too.
A: You can use $(x+x^2+x^3+x^4+x^5+x^6)^3$ and find the coefficient of $x^{14}.$ That comes out $15$ as in the answer. The full expansion also gives the numbers of ways to get any of the possible totals:
$$x^{18}+3x^{17}+6x^{16}+10x^{15}+15x^{14}+21x^{13}+ 
25x^{12}+27x^{11}+ \\27x^{10}+25x^9+21x^8+15x^7+10x^6+6x^5+3x^4+x^3.$$
Note the symmetry here: same number of ways to get $k$ as to get $21-k.$ That makes sense since a map taking $(x,y,z)$ to $(7-x,7-y,7-z)$ gives a bijection from cases with sum $x$ and cases with sum $21-x.$
As noted in my comment below, $\binom{13}{11}$ counts all cases, even those in which some of the $x_i$ exceed $5$ (which sould not happen if each die roll is written as $1+x_i$). The use of counting sums of nonnegative things to equal a given number does not have any upper bound on the summands built into it. I couldn't follow the rest of the attempted calculation, seemingly based on inclusion/exclusion or some other way of omitting all the cases where any $x_i \ge 6.$ [Note that it is $6$ or higher which should be omitted here, since $x_i$ has been defined as one less than the outome for die $i$]
Added: OP's method works in this case.
If one starts with each die having $x_i+1$ as a result, then the dice sum $14$ does give $x_1+x_2+x_3=11$ which for nonnegative $x_i$ has $\binom{13}{2}=78$ solutions. To count the number of these which are "bad" in that they have one of the $x_i \ge 6$ is made easier in this case by the fact that if one $x_i \ge 6$ the other two are not, so that one can triple the count of solutions to $y_1+y_2+y_3=11-6=5,$ which is $\binom{7}{2}=21.$ Thus we get $78-3\cdot 21=15$ for the final count.
A: The OP asked where he or she went wrong, so I'm only going to address that.  The basic problem is that the property $P_i$ should have been $x_i\ge6$ instead of $x_i\ge7$.  Using the correct property leads to $\omega(P_i)={(11-6)+3-1\choose(11-6)}={7\choose5}=21$.
The mistake stemmed from using the same variables in the transformed equation and then using the inequality restrictions from the original equation.  If you had rewritten the new equation as $(u_1+1)+(u_2+1)+(u_3+1)=14$ with $0\le u_i\le5$, it might have been more clear that you need to remove solutions with $u_i\ge6$.
