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: \[ \text{Valid solutions} = 78 - 21 = 57. \]

Answer 2

To solve the problem of distributing 15 apples to persons A, B, and C with specific constraints, we can follow these steps:

1. First, satisfy the requirement that A and C each receive at least 2 apples. This uses up a total of \(2 + 2 = 4\) apples, leaving us with \(15 - 4 = 11\) apples to distribute freely among A, B, and C.

2. Let \(a\), \(b\), and \(c\) represent the number of remaining apples given to A, B, and C, respectively. Then we have the equation:

\(a + b + c = 11\)

3. Now, apply the constraint that B can receive at most 5 apples, implying:

\(0 \leq b \leq 5\)

4. We must compute the valid combinations for each possible value of \(b\) from 0 to 5:

1. For \(b = 0\):

   \(a + c = 11\)

   Number of solutions: \(\binom{12}{1} = 12\)
2. For \(b = 1\):

   \(a + c = 10\)

   Number of solutions: \(\binom{11}{1} = 11\)
3. For \(b = 2\):

   \(a + c = 9\)

   Number of solutions: \(\binom{10}{1} = 10\)
4. For \(b = 3\):

   \(a + c = 8\)

   Number of solutions: \(\binom{9}{1} = 9\)
5. For \(b = 4\):

   \(a + c = 7\)

   Number of solutions: \(\binom{8}{1} = 8\)
6. For \(b = 5\):

   \(a + c = 6\)

   Number of solutions: \(\binom{7}{1} = 7\)

5. Sum these solutions to find the total number of distributions:

\(12 + 11 + 10 + 9 + 8 + 7 = 57\)

Hence, the number of ways to distribute 15 apples under the given conditions is \(57\).