In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins

Question

In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively

Answer 1

Let $a,b$, and $c$ be the numbers of coins donated by A, B, and C, respectively. Then you want the number of solutions in non-negative integers to the equation $a+b+c=10$, subject to the condition that $a\le 6,b\le 7$, and $c\le 8$. This is a standard stars-and-bars problem with upper bounds. Without the upper bounds there would be $$\binom{10+3-1}{3-1}=\binom{12}2$$ solutions; the linked article gives a pretty good explanation of this calculation.

Now you have to subtract the impossible solutions, i.e., those that would have someone donating more money than he has. In this problem it’s not hard, because it’s impossible for more than one of $a,b$, and $c$ to exceed its upper limit and still have a total of $10$. Thus, you need to subtract the number of solutions with $a\ge 7$, the number with $b\ge 8$, and the number with $c\ge 9$.

Note that if you assume that $a\ge 7$, you might as well replace it by $a'=a-7$ and count non-negative solutions of $a'+b+c=3$: every one of those corresponds to exactly one solution of $a+b+c=10$ with $a\ge 7$, and vice versa. Counting solutions of $a'+b+c=3$ in non-negative integers is just the basic stars-and-bars problem again, this time without constraints, so you can use the formula from the linked article without any modification.

The cases $b\ge 8$ and $c\ge 9$ can be handled similarly.

Answer 2

Let A,B,C donate $x_1,x_2,x_3$ coins with $x_i\geq 0$.

Then A/Q $\sum x_i=10$ .....(1)

At first lets find all the solutions of this equation in integers.

The no. of such solution is $12C2$

Now we will find the no. of solution in which $x_1\geq 7$,(these solutions cant be considered),

To find this lets replace $x_1$ by $x+6$ where $x\geq 1$ putting this into 1 we have $x+x_2+x_3=4$ we will find no. of such solns. $5C2$

In this way we will find the other cases which are not possible.

Namely when $x_2\geq8$ in this case we have $4C2$ solutions , and the last one when $x_3\geq 9$ then we have $3C2$ solutions.

All the cases which cant be posiible are disjoint implying the total no. of solution =(total no. of cases)-(cases not possible).

$12C2-5C2-4C2-3C2$