In how many different ways can three persons | vedclass

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(A) Let $x_1, x_2, x_3$ be the number of coins donated by $A, B, C$ respectively. We need to find the number of non-negative integer solutions to $x_1

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(A) Let $x_1, x_2, x_3$ be the number of coins donated by $A, B, C$ respectively. We need to find the number of non-negative integer solutions to $x_1 + x_2 + x_3 = 10$ subject to $0 \le x_1 \le 6, 0 \le x_2 \le 7, 0 \le x_3 \le 8$.Total solutions without constraints using stars and bars formula ${ }^{(n+r-1)} C_{r-1}$ is ${ }^{(10+3-1)} C_{3-1} = { }^{12} C_2 = 66$.Using the Principle of Inclusion-Exclusion:Let $S$ be the set of all non-negative solutions $(|S| = 66)$.Let $P_1$ be the condition $x_1 \ge 7$,$P_2$ be $x_2 \ge 8$,$P_3$ be $x_3 \ge 9$.Number of solutions satisfying $P_1$: $x_1' + x_2 + x_3 = 10 - 7 = 3 \implies { }^{(3+3-1)} C_2 = { }^5 C_2 = 10$.Number of solutions satisfying $P_2$: $x_1 + x_2' + x_3 = 10 - 8 = 2 \implies { }^{(2+3-1)} C_2 = { }^4 C_2 = 6$.Number of solutions satisfying $P_3$: $x_1 + x_2 + x_3' = 10 - 9 = 1 \implies { }^{(1+3-1)} C_2 = { }^3 C_2 = 3$.Intersections like $P_1 \cap P_2$: $x_1 + x_2 + x_3 = 10 - 7 - 8 = -5$ (Impossible,$0$ ways).Total valid ways $= 66 - (10 + 6 + 3) + 0 = 47$.

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

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