Let N be the number of ways in which 11 identical balls can be distributed among three boys. If each receives at least 3, then N is?
Understand the Problem
The question is asking for the number of ways to distribute 11 identical balls among three boys, with the condition that each boy must receive at least 3 balls. To solve it, we can use the concept of combinatorial counting methods.
Answer
The number of ways is 6.
Answer for screen readers
The number of ways to distribute the 11 identical balls among the three boys, ensuring each receives at least 3 balls, is 6.
Steps to Solve
- Set up the problem with conditions
We have 11 identical balls and 3 boys. Since each boy must receive at least 3 balls, we can start by giving 3 balls to each boy.
- Calculate the remaining balls
After giving 3 balls to each boy, we have:
$$ 11 - (3 \times 3) = 11 - 9 = 2 $$
So, we have 2 balls left to distribute.
- Apply the stars and bars theorem
Now we need to distribute these 2 remaining balls among the 3 boys. Using the stars and bars combinatorial method, the formula to find the number of ways to distribute $n$ identical items into $k$ distinct groups is given by:
$$ \binom{n + k - 1}{k - 1} $$
In our case, $n = 2$ (remaining balls) and $k = 3$ (boys).
- Calculate the combinations
Substituting the values into the formula:
$$ \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} $$
- Compute the binomial coefficient
Calculate:
$$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 $$
Thus, there are 6 ways to distribute the remaining balls.
The number of ways to distribute the 11 identical balls among the three boys, ensuring each receives at least 3 balls, is 6.
More Information
This problem applies the "stars and bars" theorem, which is a common technique in combinatorial mathematics used to solve problems of distributing indistinguishable objects into distinguishable bins.
Tips
- Forgetting to initially give each boy the required minimum number of balls.
- Miscalculating the binomial coefficient, particularly in factorial calculations.
- Not applying the “stars and bars” theorem correctly, confusing the number of stars (balls) and bars (dividers).
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