Question

In how many ways can 40 sweets be given to A, B, C, and D such that:
- B gets at least 3 sweets,
- D gets at least 5 sweets,
- A and C may get zero sweets.

• A. 4960
• B. 6545
• C. 9139
• D. 15

Hint:

When there are minimum conditions, allocate them first, then solve the unrestricted distribution using stars-and-bars.

Answer 1

The Correct Option is B

Step 1: Adjust for minimums

Give \( B \) 3 sweets and \( D \) 5 sweets initially to satisfy their minimum requirements. The sweets remaining are: \[ 40 - (3 + 5) = 32 \]

Step 2: Distribute without restriction

Now, distribute the remaining 32 sweets among 4 people (\( A, B, C, D \)) without any restrictions. This is a stars-and-bars problem: \[ \binom{32 + 4 - 1}{4 - 1} = \binom{35}{3} \]

Step 3: Calculate

\[ \binom{35}{3} = \frac{35 \cdot 34 \cdot 33}{6} = 6545 \]

Final Answer:

\[ \boxed{6545} \]