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Chapter 5: Problem 425

The number of ways of distributing 8 identical balls in 3 distinct boxes so that no box is empty is (a) 5 (b) \({ }^{8} \mathrm{C}_{3}\) (c) 38 (d) 21

Short Answer

Expert verified
There are 21 ways to distribute 8 identical balls in 3 distinct boxes, ensuring that no box is empty. The correct answer is (d) 21. We used the "stars and bars" combinatorial method to calculate this by first putting one ball in each box and then arranging the remaining 5 balls and 2 bars to represent the distribution among the boxes. The number of ways to arrange the balls and bars was calculated using the formula \( \frac{(\text{total objects})!}{(\text{count of balls})!(\text{count of bars})!} \) and simplifying the expression.

Step by step solution

01

Understanding the "Stars and Bars" Method

The "stars and bars" method is a combinatorial technique used to count the number of ways to distribute objects (in this case, balls) among a certain number of containers (in this case, boxes). In this method, imagine the balls as stars, and we will put bars between them to form "partitions" that indicate the division of balls among the boxes. The method will help to ensure no box is empty.
02

Ensure No Box is Empty

Since no box can be empty, let's start by putting one ball in each box first. So, we are now left with 5 balls that need to be distributed among the 3 distinct boxes.
03

Arrange the remaining balls and bars

We have 5 balls left to distribute among the 3 boxes. To use the "stars and bars" method, we are going to place 2 bars (3 boxes - 1) among the 5 balls to make partitions representing each box. So, we have 5 balls (or stars) and 2 bars that we need to arrange in such a manner that the balls are separated by the bars. For example, the arrangement: * * * | * | * * represents box 1 with 3 balls, box 2 with 1 ball, and box 3 with 2 balls.
04

Calculate the number of ways to arrange balls and bars

Now, we need to calculate the number of ways to arrange these 5 balls and 2 bars. This can be found using the formula: \( \text{Ways to arrange} = \frac{(\text{total objects})!}{(\text{count of balls})!(\text{count of bars})!} \) Total objects in our case are 5 balls and 2 bars, so we have a total of 7 objects. Substituting the values: \( \text{Ways to arrange} = \frac{7!}{5!2!} \)
05

Simplify the expression

We simplify the expression to get the answer: \( \text{Ways to arrange} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)(2 \times 1)} \) By canceling out the common terms, we get: \( \text{Ways to arrange} = \frac{7 \times 6}{2} = 7 \times 3 = 21\) So, there are 21 ways to distribute 8 identical balls in 3 distinct boxes, ensuring that no box is empty. The correct answer is (d) 21.

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Most popular questions from this chapter

A man has 7 relative, 4 of them ladies and 3 gentlemen. His wife also have 7 relatives. 3 of them ladies and 4 gentlemen, They invite for a dinner partly 3 ladies and 3 gentlemen so that there are 3 of the men's relative and 3 of the wife's relative. The number of ways of invitation is (a) 854 (b) 585 (c) 485 (d) 548

The number of ways in which in a necklace can be formed by using 5 identical red beads and 6 identical black beads is (a) \([(11 !) /(6 ! 4 !)]\) (b) \({ }^{11} \mathrm{P}_{6}\) (c) \([(10 !) /(2(6 ! 5 !)]\) (d) None of these

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The number of 10 letter codes that can be formed using the characters \(\mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{r}\) with the restriction that \(\mathrm{x}\) appears exactly thrice and \(\mathrm{y}\) appears exactly twice in each such codes is (a) 60840 (b) 88400 (c) 80640 (d) 64080
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