Question

The number of ways in which 15 identical gold coins can be distributed among 3 persons such that each one gets at least 3 gold coins is

• A. \(27\)
• B. \(28\)
• C. \(22\)
• D. \(25\)

Hint:

When distributing identical objects with minimum conditions, redefine variables and use the stars and bars method.

Answer 1

The Correct Option is B

To solve the problem of distributing 15 identical gold coins among 3 persons such that each person gets at least 3 gold coins, we can use the stars and bars method with constraints. Step 1: Apply the constraints
Each person must receive at least 3 gold coins. Let:
\( x_1 \) = number of coins for person 1,
\( x_2 \) = number of coins for person 2,
\( x_3 \) = number of coins for person 3.
The constraints are: \[ x_1 \geq 3, \quad x_2 \geq 3, \quad x_3 \geq 3. \] To simplify, let: \[ y_1 = x_1 - 3, \quad y_2 = x_2 - 3, \quad y_3 = x_3 - 3. \] Now, \( y_1, y_2, y_3 \geq 0 \), and the total number of coins becomes: \[ y_1 + y_2 + y_3 = 15 - (3 + 3 + 3) = 6. \] Step 2: Use the stars and bars method We need to find the number of non-negative integer solutions to: \[ y_1 + y_2 + y_3 = 6. \] The formula for the number of non-negative integer solutions to \( y_1 + y_2 + \dots + y_k = n \) is: \[ \binom{n + k - 1}{k - 1}. \] Step 3: Applying the stars and bars method.
The number of non-negative integer solutions is given by the formula: \[ \text{Total ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28 \]