Permutation and Combination problem : In how many ways can Rs. 16 be divided into 4 person when none of them get ... Permutation and Combination problem : 
Problem : In how many ways can Rs. 16 be divided into 4 person when none of them get less than Rs. 3 ? 
I have this : The number of ways of distributing $n$ things all alike into $r$ different groups is $^{n+r-1}C_{r-1}$ lots , no lots being blank: 
Coefficient of $x^n $ in $(1+x+x^2+.....)^r$
I am unable to understand how to use this case in the given problem. Request you to please help me on this and elaborate a bit .. will greatful to you... Thanks...
 A: The easiest way to solve it is to imagine giving each person Rs. $3$ right away and then distributing the remaining Rs. $4$ arbitrarily. In other words, reduce the problem to distributing Rs. $4$ amongst $4$ people with no restrictions: you have the formula for that, with $n=r=4$:
$$\binom{4+4-1}{4-1}=\binom73=35\;.$$
A: What you want is, how many ways are there to write
$$\{1, \ldots, 16\} = A_1 \dot{\cup} A_2 \dot\cup A_3 \dot\cup A_4$$
Under the constraint, that $|A_i| \geq 3 \quad \forall\ i$.
Did I understand that correctly?
A: If we expand $$(1+x+x^2+\cdots+x^{16})(1+x+x^2+\cdots+x^{16}) \times (1+x+x^2+\cdots+x^{16})(1+x+x^2+\cdots+x^{16})$$ the coefficient of $x^{16}$ receives contributions from terms $x^a x^b x^c x^d$ where $a+b+c+d=16$ and $a,b,c,d \in \{0,1,\ldots,16\}$.
We can think of $a$ as being the amount of Rs that the first person gets, $b$ as the the amount of Rs that the second person gets, and so on.
However, the above expression admits the possibility that e.g. $a<3$, so we'd need to modify the expression to account for this restriction.
