The total number of ways in which 5 balls of | vedclass

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(B) To distribute $5$ distinct balls among $3$ distinct persons such that each person receives at least one ball,we use the principle of inclusion-exc

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(B) To distribute $5$ distinct balls among $3$ distinct persons such that each person receives at least one ball,we use the principle of inclusion-exclusion or partition the balls into groups of sizes $(3, 1, 1)$ and $(2, 2, 1)$.Case $1$: Group sizes $(3, 1, 1)$.The number of ways to partition $5$ distinct balls into groups of sizes $3, 1, 1$ is $\frac{5!}{3!1!1! \cdot 2!} = \frac{120}{6 \cdot 2} = 10$.Since the persons are distinct,we multiply by $3!$ to distribute these groups: $10 	imes 6 = 60$.Case $2$: Group sizes $(2, 2, 1)$.The number of ways to partition $5$ distinct balls into groups of sizes $2, 2, 1$ is $\frac{5!}{2!2!1! \cdot 2!} = \frac{120}{4 \cdot 2} = 15$.Since the persons are distinct,we multiply by $3!$ to distribute these groups: $15 	imes 6 = 90$.Total ways $= 60 + 90 = 150$.

The total number of ways in which $5$ balls of different colours can be distributed among $3$ persons so that each person gets at least one ball is

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