In how many different ways can 21 identical balls be distributed among 3 persons such that each person receives at least 5 balls?

Introduction / Context:
This question involves distributing identical objects among distinct persons subject to a minimum requirement. It is a standard stars and bars type problem from permutations and combinations, where the objects are identical balls and the people are distinct recipients. The minimum condition makes the problem slightly more interesting than a basic non negative integer solution count.

Given Data / Assumptions:

• There are 21 balls, all identical.
• There are 3 persons, all distinct.
• Each person must receive at least 5 balls.
• No upper limit is given besides the total of 21 balls.

Concept / Approach:
The usual method is to convert the at least 5 constraint into a non negative constraint by a simple substitution. If each person must receive at least 5 balls, we can give 5 balls to each person first, and then distribute the remaining balls without a lower bound. After this transformation, the problem reduces to finding the number of non negative integer solutions of a linear equation, which is solved with the stars and bars formula.

Step-by-Step Solution:
Step 1: Let x1, x2 and x3 be the number of balls received by the three persons.Step 2: We have x1 + x2 + x3 = 21, with each xi greater than or equal to 5.Step 3: Define new variables y1 = x1 - 5, y2 = x2 - 5, y3 = x3 - 5. Each yi is now greater than or equal to 0.Step 4: Substituting, we get (y1 + 5) + (y2 + 5) + (y3 + 5) = 21, so y1 + y2 + y3 + 15 = 21.Step 5: Simplify to get y1 + y2 + y3 = 6, with yi greater than or equal to 0.Step 6: The number of non negative integer solutions of y1 + y2 + y3 = 6 is given by (6 + 3 - 1)C(3 - 1) = 8C2.Step 7: Compute 8C2 = 8 * 7 / (2 * 1) = 28.

Verification / Alternative check:
You can verify the count by listing some sample solutions. For example, when y1 = 0, y2 + y3 = 6, which has 7 solutions (including 0,6 and 6,0). When y1 = 1, there are 6 solutions, and so on. Summing these manually produces 28 solutions and matches the stars and bars formula. This confirms that there are 28 valid distributions of the original 21 balls with each person getting at least 5.

Why Other Options Are Wrong:

• 14 and 21: These numbers can appear if someone forgets to convert to the non negative form correctly or misapplies the combination formula.
• 7: This is far too small and could result from only counting distributions with one person fixed and ignoring others.

Common Pitfalls:
A typical mistake is to directly apply the formula (21 + 3 - 1)C(3 - 1) without adjusting for the minimum of 5, which would ignore the lower bound requirement. Another mistake is to treat the balls as distinct, which is incorrect for this problem. Also, some learners only distribute the remaining balls to two people instead of three after the transformation, which changes the equation and the answer.

Final Answer:
The balls can be distributed among the three persons in 28 different ways.