Distributing Distinct Balls Among Persons with At Least One Each

To find the total number of ways to distribute $5$ distinct balls among $3$ distinct individuals such that each individual receives at least one ball, we can use two common combinatorial methods: the Principle of Inclusion-Exclusion or Grouping/Partitioning.

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Method 1: Principle of Inclusion-Exclusion (PIE)

1. Total Distributions Without Restriction: Each of the $5$ distinct balls can be given to any of the $3$ individuals. $\text{Total ways} = 3^5 = 243$

2. Subtract Distributions Where At Least One Person Receives No Balls:

   • Ways where a specific person receives no balls: Choose $1$ person to receive $0$ balls in $\binom{3}{1}$ ways. The $5$ balls are then distributed among the remaining $2$ persons in $2^5$ ways. $\binom{3}{1} \times 2^5 = 3 \times 32 = 96$

   • Ways where specific two persons receive no balls (over-subtracted): Choose $2$ persons to receive $0$ balls in $\binom{3}{2}$ ways. The $5$ balls are then distributed to the remaining $1$ person in $1^5$ way. $\binom{3}{2} \times 1^5 = 3 \times 1 = 3$

   • Ways where all three persons receive no balls: $\binom{3}{3} \times 0^5 = 0$

3. Applying PIE Formula: $\text{Required Number of Ways} = 3^5 - \binom{3}{1} \cdot 2^5 + \binom{3}{2} \cdot 1^5 - \binom{3}{3} \cdot 0^5$ $\text{Required Number of Ways} = 243 - 96 + 3 - 0 = 150$

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Method 2: Grouping (Partitioning into Subsets)

Since $5$ distinct balls are distributed into $3$ non-empty groups, the possible group sizes are:

1. Case 1: Group sizes $(3, 1, 1)$

   • Number of ways to divide $5$ distinct balls into groups of sizes $3, 1, 1$: $\frac{5!}{3! \cdot 1! \cdot 1! \cdot 2!} = \frac{120}{6 \cdot 2} = 10$
   • Distributing these $3$ groups to $3$ distinct individuals: $10 \times 3! = 10 \times 6 = 60\text{ ways}$

2. Case 2: Group sizes $(2, 2, 1)$

   • Number of ways to divide $5$ distinct balls into groups of sizes $2, 2, 1$: $\frac{5!}{2! \cdot 2! \cdot 1! \cdot 2!} = \frac{120}{2 \cdot 2 \cdot 2} = 15$
   • Distributing these $3$ groups to $3$ distinct individuals: $15 \times 3! = 15 \times 6 = 90\text{ ways}$

3. Total Ways: $\text{Total ways} = 60 + 90 = 150$

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Both methods consistently yield $150$.

Thus, the correct option is (B).