Question

The number of ways 16 oranges distributed to 4 children, each gets at least one.

• A. 403
• B. 384
• C. 429
• D. 455

Hint:

Stars and bars method: For $x_1+...+x_r=n, x_i \ge 1$, ways = $\binom{n-1}{r-1}$.

Answer 1

The Correct Option is D

To solve the problem of distributing 16 oranges to 4 children such that each child receives at least one orange, we need to use the "stars and bars" theorem in combinatorics.

Here’s a step-by-step explanation for solving this problem:

1. Since each child must receive at least one orange, assign one orange to each child initially. This accounts for 4 oranges, leaving us with 16 - 4 = 12 oranges to distribute freely.
2. Now, the problem reduces to finding the number of ways to distribute these remaining 12 oranges among the 4 children. This can be framed as finding the number of non-negative integer solutions to the equation: x_1 + x_2 + x_3 + x_4 = 12
3. According to the stars and bars method, the number of such solutions is given by the formula: \binom{n+k-1}{k-1}, where n is the total number of items to distribute (12 in this case) and k is the number of recipients (4 in this case).
4. Plugging in the values, we have: \binom{12+4-1}{4-1} = \binom{15}{3}
5. Calculate \binom{15}{3}:
   • The formula for combination is: \binom{n}{r} = \frac{n!}{r!(n-r)!}
   • Applying this to our numbers: \binom{15}{3} = \frac{15!}{3! \cdot 12!}
   • Simplifying: \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455

Thus, the correct answer is 455, which matches the given correct answer option.