Looking for Distinct Solutions to $x_1 + x_2 + x_3 + x_4 = 100$ Given Certain Conditions How many distinct solutions to the equation does the following have?
$$x_1 + x_2 + x_3 + x_4  = 100$$ such that $x_1 \in \{0,1,2,... 10 \}, x_2, x_3, x_4 \in \{0,1,2,3,...\}$
My attempt: Ordinarily if the range of all the $x_i$s were the set of natural numbers including zero, I could just do $_{103}C_3$. However since $x_1$ is bound from $0$ to $10$ inclusive, I can't do that. What I attempted was to set $x_1$ to each value from $0$ to $10$ and then take the summation of all the distinct solutions for each x_1. 
Hence,
$$\sum^{10}_{i=0}(_{102-i}C_2)$$
So for each value of $x_1$ I find how many distinct solutions to the equation 
$$x_2 + x_3 + x_4  = 100 - x_1$$
there are and then sum them together. 
Is this the right approach? It feels kinda awkward to me? Is there a more efficient option?
 A: Your attempt is correct. You just need to do:
$$x_2 + x_3 + x_4 = 100-x_1$$
So, we have:
$$\sum_{x_1 = 0}^{10} \binom{102-x_1}{2}$$
We know that:
$$\sum_{k=m}^{n} \binom{k}{m} = \binom{n+1}{m+1}$$
Hence:
$$\sum_{x_1 = 0}^{10} \binom{102-x_1}{2} = \sum_{k=2}^{102} \binom{k}{2} - \sum_{k=2}^{91} \binom{k}{2} = \binom{103}{3} - \binom{92}{3}$$
A: We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 = 100 \tag{1}$$
in the nonnegative integers subject to the restriction that $x_1 \leq 10$.
Since a particular solution of equation 1 corresponds to the placement of $4 - 1 = 3$ addition signs in a row of $100$ ones, there are 
$$\binom{100 + 4 - 1}{4 - 1} = \binom{103}{3}$$
solutions of the equation in the nonnegative integers since we must choose which three of the $103$ positions required for $100$ ones and three addition signs will be filled with addition signs.
From these, we must subtract those solutions in which $x_1 > 10$.  
Let's count the number of solutions that violate the restriction that $x_1 \leq 10$.  Suppose $x_1 > 10$.  Let $x_1' = x_1 - 11$.  Then $x_1'$ is a nonnegative integer.  Substituting $x_1' + 11$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 1 + x_2 + x_3 + x_4 & = 100\\
x_1' + x_2 + x_3 + x_4 & = 89 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{89 + 4 - 1}{4 - 1} = \binom{92}{3}$$
solutions.
Hence, the number of nonnegative integer solutions of equation 1 that satisfy the restriction that $x_1 \leq 10$ is 
$$\binom{103}{3} - \binom{92}{3}$$
A: You can use the $k$-combination with repetitions. When I write as $_{n}H_{k}=\mathrm{}_{n+k-1}C_k$ instead of $\left(\!\!\binom{n}{k}\!\!\right)=\binom{n+k-1}{k}$, the following is true:
$$\begin{align}\frac{(n+k)!}{k!n!}&=\mathrm{}_{n+k}C_{k}=\mathrm{}_{n+1}H_k\\
&=\mathrm{}_n H_k+\mathrm{}_{n+1}H_{k-1}\\
&=\mathrm{}_{n}H_k+\mathrm{}_{n}H_{k-1}+\cdots+\mathrm{}_{n}H_1+\mathrm{}_{n+1}H_0.\end{align}$$
Therefore, the result you are looking for is $\mathrm{}_4H_{100}-\mathrm{}_4H_{89}$.
