Solve for $x$ and $y$: $\sum_{n=3}^{n=10} {^nC_3}\cdot{^{14-n}C_4}={^xC_y}$. $$^3C_3\cdot{^{11}C}_4+^4C_3\cdot{^{10}C}_4+^5C_3\cdot{^9C}_4+\cdots+{^{10}C}_3\cdot{^4C}_4= {^xC_y}$$ where $y$ is an odd natural number.
My Thinking: I basically thought of a very foolish idea. Basically, it's looking like these are number of ways of some particular case. I thought that if we have a $14$ players, then these are the number of ways of dividing them into two teams $A$ and $B$ and then selecting players from them as per the condition that we have to select at least three players from Team $A$ and at least four players from Team $B$. But it doesn't provide any appropriate result and I know I messed something here. Further as a second method, $$\sum_{n=3}^{n=10} {^nC_3}\cdot{^{14-n}C_4}$$ This method too is very lengthy and not helpful. If anyone can spot, what actually the terms of this series are or any alternate solutions, it would be a great help. Thank You
 A: Let $S=\{1,2,3, \ldots ,15\}$. Suppose we want to count subsets of size $8$ of this set. Then it would be $\binom{15}{8}=\color{red}{\binom{15}{7}}$.
But we can count this in another way as follows:
Suppose we agree to write each subset in increasing order of elements, then in each subset of size $8$ let us focus on the fourth smallest element in that subset, e.g. $\{1,2,3,\color{red}{4},5,6,7,8\}$. Note that the $4$ is the smallest possible value for this element. In that case we have $\binom{3}{3}$ choices of elements below it and $\binom{11}{4}$ choices of elements above it (because you will have to pick elements from $\{5,6,7,\ldots, 15\}$ . In all, that will give us $\color{blue}{\binom{3}{3}\binom{11}{4}}$ subsets of size $8$, where $4$ is the fourth smallest element in the subset.
Now we do the same when $5$ is the fourth smallest element. We can choose three from $\{1,2,3,4\}$ and four from $\{6,7,8, \ldots, 15\}$. This can be done in $\color{blue}{\binom{4}{3}\binom{10}{4}}$ ways.
If we continue like this then the last case will be when $11$ is the fourth smallest element in the subset. This can be done in $\color{blue}{\binom{10}{3}\binom{4}{4}}$ ways.
Since the two ways of counting answer the same question, we have
$$\color{magenta}{\sum_{n=3}^{10}\binom{n}{3}\binom{14-n}{4}=\binom{15}{7}}.$$
