<s>[INST] Number of ways 6 things can be selected from a group of identical things 
In how many ways can a party of 6 men be selected out of 10 Hindus, 8 Muslims, and 6 Christians if the party consists of at least one person of each religion( Consider only the religion of the person?

Now, I know this problem can be solved by finding the non-negative integral solution to the given equation,
$$x+y+z=6$$
Which the answer turns out to be $\binom{6-1}{3-1}$
But I want to know is there any other method of solving this problem? 
Any help would be appreciated.
 [/INST] Stars and Bars
Once we pre-select one of each religion, this is the same as asking "how many ways we can choose $3$ men from $9$ Hindus, $7$ Muslims, and $5$ Christians?" This is the same as asking how many ways to solve
$$
x+y+z=3
$$
with $x,y,z\ge0$. Stars and Bars say $\binom{5}{2}=10$.

Generating Functions
$$
\begin{align}
&\left[x^n\right]
\overbrace{\left(x+\cdots+x^{10}\right)}^{1\dots10\text{ Hindus}}
\overbrace{\left(x+\cdots+x^8\right)}^{1\dots8\text{ Muslims}}
\overbrace{\left(x+\cdots+x^6\right)}^{1\dots6\text{ Christians}}\\
&=\left[x^n\right]x^3\frac{1-x^{10}}{1-x}\frac{1-x^8}{1-x}\frac{1-x^6}{1-x}\\
&=\left[x^{n-3}\right]\frac{1-x^6-x^8-x^{10}+x^{14}+x^{16}+x^{18}-x^{24}}{(1-x)^3}\\
&=\left(\left[x^{n-3}\right]-\left[x^{n-9}\right]-\left[x^{n-11}\right]-\left[x^{n-13}\right]+\left[x^{n-17}\right]+\left[x^{n-19}\right]+\left[x^{n-21}\right]-\left[x^{n-27}\right]\right)\frac1{(1-x)^3}\\
&=\textstyle(-1)^{n-3}\left[\binom{-3}{n-3}-\binom{-3}{n-9}-\binom{-3}{n-11}-\binom{-3}{n-13}+\binom{-3}{n-17}+\binom{-3}{n-19}+\binom{-3}{n-21}-\binom{-3}{n-27}\right]\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{\textstyle\binom{n-1}{n-3}-\binom{n-7}{n-9}-\binom{n-9}{n-11}-\binom{n-11}{n-13}+\binom{n-15}{n-17}+\binom{n-17}{n-19}+\binom{n-19}{n-21}-\binom{n-25}{n-27}}
\end{align}
$$
This is the number of ways to pick a group of $n$ people from a pool of $10$ Hindus, $8$ Muslims, and $6$ Christians, where there is at least $1$ of each.
The only one of these that is not $0$ for $n=6$ is $\binom{n-1}{n-3}=\binom{5}{3}$.
Note that we could reduce the last line to
$$
\textstyle\binom{n-1}{2}-\binom{n-7}{2}-\binom{n-9}{2}-\binom{n-11}{2}+\binom{n-15}{2}+\binom{n-17}{2}+\binom{n-19}{2}-\binom{n-25}{2}
$$
except that the equation $\binom{n}{k}=\binom{n}{n-k}$ only holds when $k$ and $n-k$ are non-negative integers.
</s>