Let $a,b,c$ be whole numbers so that $1 \le a \le 3$, $2\le b \le4$, $3\le c \le5$. Find number of solutions of the equation, $a+b+c=10$. Let $a,b,c$ be whole numbers so that $$1 \le a \le 3$$ $$2\le b \le4$$ $$3\le c \le5$$Find number of solutions of the equation, $a+b+c=10$. Note: Please use FPC/PnC to answer this instead of the binomial theorem
My Approach:
I treated this question as a variant of the normal Beggar's Method problem. so a,b,c are beggars and I need to distribute 10 coins among them. Since all the beggar's need to have a minimum amount of coins, I 'give' them those so the new question is, distributing 4 coins among 3 beggars such that:
$$a \le 3$$ $$b \le4$$ $$c \le5$$
I couldn't solve this system of inequalities.
beggar's method
 A: That is correct. We need to find number of solutions to,
$A + B + C = 4, $ where $A, B, C$ are non-negative numbers and $A \leq 2, B \leq 2, C \leq 2$. You can easily count case by case.
Alternatively, we could rewrite it as $A = 2 - x, B = 2 - y, C = 2 - z$ and find number of solutions to
$x + y + z = 2, x \leq 2, y \leq 2, z \leq 2$
Using stars and bars, we get ${4 \choose 2} = 6$ solutions.
Now coming back to counting number of solutions to $A + B + C = 4$, please see the cases below. Can you count number of solutions in each case?

*

*If $A = 0$, we have $B + C = 4$

*If $A = 1$, we have $B + C = 3$

*If $A = 2$, we have $B + C = 2$
A: We can define the followings:
$$
\begin{align}
x&=a\\
y&=b-1\\
z&=c-2
\end{align}
$$
Our problem is now to find the number of positive integer solutions of $x+y+z=7$ in which all three are not greater than $3$. The solution without restriction is simple stars and bars:
$$
\binom{7-1}{3-1}=15
$$
The solution with one of $x,y,z$ exceeding $3$ is the solution of:
$$
\begin{align}
x_{1}+y_{1}+z_{1}&=4\phantom{xx}x_{1},y_{1},z_{1}\in\mathbb{N}\\
x_{2}+y_{2}+z_{2}&=3\phantom{xx}x_{2},y_{2},z_{2}\in\{0,3\}\\
\\
x&=x_{1}+x_{2}\\
y&=y_{1}+y_{2}\\
z&=z_{1}+z_{2}
\end{align}
$$
The number of solutions can be calculated using stars and bars, permutation, and multiplication rule. So the answer to the original question is:
$$
\binom{7-1}{3-1}-3\cdot\binom{4-1}{3-1}=6
$$
