Question

The number of ways of distributing 15 apples to three persons A, B, C such that A and C each get at least 2 apples and B gets at most 5 apples is:

• A. \( 57 \)
• B. \( 131 \)
• C. \( 156 \)
• D. \( 251 \)

Hint:

For combinatorial distribution problems, use the stars and bars technique and apply constraints carefully.

Answer 1

The Correct Option is A

Step 1: Define the Variables Let \( x_A, x_B, x_C \) represent the number of apples received by A, B, and C respectively. The total number of apples distributed is: \[ x_A + x_B + x_C = 15. \] Given constraints: - A and C must each receive at least 2 apples: \[ x_A \geq 2, \quad x_C \geq 2. \] - B must receive at most 5 apples: \[ 0 \leq x_B \leq 5. \]
Step 2: Transform the Equation Define new variables: \[ y_A = x_A - 2, \quad y_C = x_C - 2. \] Since A and C receive at least 2 apples, these new variables \( y_A \) and \( y_C \) can take non-negative values. Thus, the equation becomes: \[ y_A + x_B + y_C = 15 - (2 + 2) = 11. \]
Step 3: Count the Ways Without Constraint on \( x_B \) Ignoring the upper bound on \( x_B \), the number of solutions to: \[ y_A + x_B + y_C = 11 \] in non-negative integers is given by the stars and bars method: \[ \text{Total solutions} = \binom{11+2}{2} = \binom{13}{2} = \frac{13 \times 12}{2} = 78. \]
Step 4: Apply Constraint on \( x_B \) Since \( x_B \leq 5 \), we must exclude cases where \( x_B \geq 6 \). Substituting \( x_B = 6 + k \) where \( k \geq 0 \), the equation transforms into: \[ y_A + k + y_C = 5. \] The number of solutions is: \[ \binom{5+2}{2} = \binom{7}{2} = \frac{7 \times 6}{2} = 21. \]
Step 5: Compute the Final Count Using the Inclusion-Exclusion Principle: \[ \text{Valid solutions} = 78 - 21 = 57. \]