Combinatorics. Please explain me how to do it. In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!

Let $a,b,c$ be the amounts that $A,B$ and $C$ give resp. Then we want to count the number of distinct solutions to $a + b + c = 10$ under the condition that $a,b,c$ are positive integers and $a \le 6, b \le 7, c \le 8$. 
One way to compute this is to compute the coefficient of $x^{10}$ in the expression $(1+x+\ldots +x^6)(1+x+\ldots +x^7)(1+x+\ldots +x^8)$ ($a$ is the exponent we choose in the first term, $b$ in the second etc., so the fact that we go up to $x^6$ in the first term expresses the $a \le 6$ and so on.) 
We can write these terms as $\frac{1-x^7}{1-x}$, $\frac{1-x^8}{1-x}$ and $\frac{1-x^9}{1-x}$, respectively, so this product equals 
$$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$ and then we can use the general Newton formula to compute the coefficient of $x^{10}$ (we expand $(1-x)^{-3}$ using that, and then count the (not too many) ways the first terms give rise to a power of $x$ that is $\le 10$).
Expanded: the general binomial implies 
$$(1-x)^{-3} = \sum_{n=0}^{\infty} \binom{k+2}{2} x^k$$
e.g. see here, and now note that we can form $x^{10}$ by picking 1's in the first 3 terms and the coefficient of $x^{10}$ in this expansion, so $\binom{12}{2}$, and also by picking $-x^7$,1,1 and $\binom{5}{3}$ (term for $x^3$), $1,-x^8,1$ and the $x^2$ term and finally $1,1,-x^9$ and the term for $x^1$ in the infinite expansion.
So we get $$\binom{12}{2} - \binom{5}{3} - \binom{4}{2} - \binom{3}{1} = 47$$
