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5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )?
Most Upvoted Answer
5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Nu...
Is it 21 ways? this question is same as find the total number of positive integral solutions of x+y+z=5 where x ,y & z are the the number of balls in each boxes there is a formula for total number of positive integral solutions (n+r-1)C(r-1) so put n=5 and r=3. see if the answer is correct otherwise there are different methods
Community Answer
5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Nu...
Understanding the Problem
To place 5 identical balls into 3 different boxes such that no box remains empty, we can use the stars and bars combinatorial method. However, since each box must contain at least one ball, we first allocate one ball to each box.
Step 1: Initial Allocation
- We start by placing 1 ball in each of the 3 boxes.
- This consumes 3 balls, leaving us with 2 balls to distribute.
Step 2: Distributing Remaining Balls
- Now, we need to distribute the remaining 2 identical balls into the 3 boxes.
- This scenario can now be solved using the stars and bars theorem.
Application of Stars and Bars
- The formula for distributing \( n \) identical items into \( k \) distinct boxes is given by:
\[
\text{Number of ways} = \binom{n + k - 1}{k - 1}
\]
- Here, \( n = 2 \) (remaining balls) and \( k = 3 \) (boxes).
Calculation
- Plugging in the values:
\[
\text{Number of ways} = \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2}
\]
- Calculating \( \binom{4}{2} \):
\[
\binom{4}{2} = \frac{4!}{2! \cdot (4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]
Final Result
- Therefore, the total number of ways to distribute 5 identical balls into 3 distinct boxes, ensuring no box is empty, is 6.
To place 5 identical balls into 3 different boxes such that no box remains empty, we can use the stars and bars combinatorial method. However, since each box must contain at least one ball, we first allocate one ball to each box.
Step 1: Initial Allocation
- We start by placing 1 ball in each of the 3 boxes.
- This consumes 3 balls, leaving us with 2 balls to distribute.
Step 2: Distributing Remaining Balls
- Now, we need to distribute the remaining 2 identical balls into the 3 boxes.
- This scenario can now be solved using the stars and bars theorem.
Application of Stars and Bars
- The formula for distributing \( n \) identical items into \( k \) distinct boxes is given by:
\[
\text{Number of ways} = \binom{n + k - 1}{k - 1}
\]
- Here, \( n = 2 \) (remaining balls) and \( k = 3 \) (boxes).
Calculation
- Plugging in the values:
\[
\text{Number of ways} = \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2}
\]
- Calculating \( \binom{4}{2} \):
\[
\binom{4}{2} = \frac{4!}{2! \cdot (4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]
Final Result
- Therefore, the total number of ways to distribute 5 identical balls into 3 distinct boxes, ensuring no box is empty, is 6.
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5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )?
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5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )?.
5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 balls are to be placwd in 3 boxes .each an hold all five 5 balls .Numbee of ways in which I ch ball can be placed so thT. no box eemaon empty ,if balls are identical but boxes are different )?.
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