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

Engineering

Mathematics

Formation of Groups

Question

210

243

150

Number of ways $= \frac{5! \times 3!}{1! 2! 2! 2!} + \frac{5! \times 3!}{1! 1! 3! 2!} = 90 + 60 = 150$

Concept: Distributing Distinct Objects with Constraints

We need to distribute 5 distinct balls (each of a different color) among 3 persons such that each person gets at least one ball. This is a problem of surjective functions (onto functions) from a set of 5 elements to a set of 3 elements, where no person is left empty-handed.

Step 1: Understand the Total Unrestricted Ways

If there were no restrictions, each of the 5 distinct balls can be given to any of the 3 persons. So, for each ball, there are 3 choices.

Total ways without restrictions: $3^{5} = 243$

Step 2: Subtract the Invalid Cases

But we must exclude cases where at least one person gets no ball. We use the principle of inclusion-exclusion.

Let:

$A$ be the set of distributions where Person 1 gets no ball.
$B$ be the set where Person 2 gets no ball.
$C$ be the set where Person 3 gets no ball.

We want the number of distributions where none of A, B, or C occurs, i.e., $N = 3^{5} - | A \cup B \cup C |$

Step 3: Apply Inclusion-Exclusion Principle

By inclusion-exclusion:

| A \cup B \cup C | = | A | + | B | + | C | - | A \cap B | - | A \cap C | - | B \cap C | + | A \cap B \cap C |

Calculate each term:

$| A | = 2^{5} = 32$ (balls go only to Persons 2 and 3)
Similarly, $| B | = 32$ , $| C | = 32$
$| A \cap B | = 1^{5} = 1$ (all balls go to Person 3 only)
Similarly, $| A \cap C | = 1$ , $| B \cap C | = 1$
$| A \cap B \cap C | = 0^{5} = 0$ (no person gets any ball, impossible for 5 balls)

So,

| A \cup B \cup C | = 32 + 32 + 32 - 1 - 1 - 1 + 0 = 96 - 3 = 93

Step 4: Compute Valid Distributions

Valid distributions where each person gets at least one ball:

N = 243 - 93 = 150

Final Answer

The total number of ways is $150$ .

Formulae and Theory

Number of onto functions from a set of m elements to a set of n elements (m ≥ n):

N = \sum_{k = 0}^{n} {(- 1)}^{k} \cdot (\binom{n}{k}) \cdot {(n - k)}^{m}

For m=5, n=3:

N = (\binom{3}{0}) 3^{5} - (\binom{3}{1}) 2^{5} + (\binom{3}{2}) 1^{5} - (\binom{3}{3}) 0^{5} = 1 \cdot 243 - 3 \cdot 32 + 3 \cdot 1 - 1 \cdot 0 = 243 - 96 + 3 = 150

Detailed Solution

Concept: Distributing Distinct Objects with Constraints

Step 1: Understand the Total Unrestricted Ways

Step 2: Subtract the Invalid Cases

Step 3: Apply Inclusion-Exclusion Principle

Step 4: Compute Valid Distributions

Final Answer

Related Topics

Formulae and Theory

Detailed Solution

Concept: Distributing Distinct Objects with Constraints

Step 1: Understand the Total Unrestricted Ways

Step 2: Subtract the Invalid Cases

Step 3: Apply Inclusion-Exclusion Principle

Step 4: Compute Valid Distributions

Final Answer

Related Topics

Formulae and Theory

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